الاخوة الاعضاء اذا كانت مشاركتك القديمة لا تظهر: اقتح تعديل المشاركة ثم استبدل com بـ net رجاءً

# الموضوع: بعض التكاملات التي لا يمكن ان تحل

1. ## رد: بعض التكاملات التي لا يمكن ان تحل

$\dpi{150}&space;Let{\color{Blue}&space;f_{2}\left&space;(&space;x&space;\right&space;)}={\color{DarkGreen}&space;\left&space;(&space;x+1&space;\right&space;)}{\color{Magenta}&space;\left&space;(&space;x^{2}-2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}{\color{Red}&space;----\left&space;(&space;9&space;\right&space;)}$

$\dpi{150}&space;{\color{Blue}&space;f_{2}\left&space;(&space;0&space;\right&space;)}={\color{DarkGreen}&space;\left&space;(&space;0+1&space;\right&space;)}{\color{Magenta}&space;\left&space;(&space;0^{2}-2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;).0+1&space;\right&space;)}={\color{Blue}&space;1}{\color{Red}&space;----\left&space;(10&space;\right&space;)}$

$\dpi{150}&space;{\color{Blue}&space;f_{2}\left&space;(&space;1&space;\right&space;)}={\color{DarkGreen}&space;\left&space;(&space;1+1&space;\right&space;)}{\color{Magenta}&space;\left&space;(&space;1^{2}&space;-2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;).1+1\right&space;)}={\color{DarkGreen}&space;2}{\color{Magenta}&space;\left&space;(&space;2-2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)&space;\right&space;)}{\color{Red}&space;---\left&space;(&space;11&space;\right&space;)}$

$\dpi{150}&space;{\color{Blue}&space;f_{2}\left&space;(&space;-1&space;\right&space;)}={\color{Blue}&space;f_{2}\left&space;(&space;e^{\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}={\color{Blue}&space;f_{2}\left&space;(&space;e^{\left&space;(&space;\frac{-3\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}={\color{Blue}&space;0}{\color{Red}&space;-----\left&space;(&space;12&space;\right&space;)}$

$\dpi{150}&space;Let{\color{Magenta}&space;f_{3}\left&space;(&space;x&space;\right&space;)}={\color{DarkGreen}&space;\left&space;(&space;x+1&space;\right&space;)}{\color{Blue}&space;\left&space;(&space;x^{2}-2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}{\color{Red}&space;----\left&space;(&space;13&space;\right&space;)}$

$\dpi{150}&space;{\color{Magenta}&space;f_{3}\left&space;(0&space;\right&space;)}={\color{DarkGreen}&space;\left&space;(&space;0+1&space;\right&space;)}{\color{Blue}&space;\left&space;(&space;0^{2}-2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;).0+1&space;\right&space;)}={\color{Magenta}&space;1}{\color{Red}&space;----\left&space;(&space;14&space;\right&space;)}$

$\dpi{150}&space;{\color{Magenta}&space;f_{3}\left&space;(1&space;\right&space;)}={\color{DarkGreen}&space;\left&space;(&space;1+1&space;\right&space;)}{\color{Blue}&space;\left&space;(&space;1^{2}-2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;).1+1&space;\right&space;)}={\color{DarkGreen}&space;2{\color{Blue}&space;\left&space;(&space;2-2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)}}{\color{Red}&space;----\left&space;(&space;15&space;\right&space;)}$

$\dpi{150}&space;{\color{Magenta}&space;f_{3}\left&space;(&space;-1&space;\right&space;)}={\color{Magenta}&space;f_{3}\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}={\color{Magenta}&space;f_{3}\left&space;(&space;e^{\left&space;(&space;\frac{-\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}={\color{Magenta}&space;0{\color{Red}&space;----\left&space;(&space;16&space;\right&space;)}}$

$\dpi{150}&space;{\color{Red}&space;\frac{x^{2}}{1+x^{5}}}={\color{Red}&space;\frac{x^{2}}{x^{5}+1}}={\color{DarkGreen}&space;\frac{a_{1}}{\left&space;(&space;x+1&space;\right&space;)}}+{\color{Blue}&space;\frac{a_{2}x+b_{2}}{\left&space;(&space;x^{2}-2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}}+{\color{Magenta}&space;\frac{a_{3}x+b_{3}}{\left&space;(&space;x^{2}-2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)+1&space;\right&space;)}}{\color{Red}&space;--\left&space;(&space;17&space;\right&space;)}$

$\dpi{150}&space;{\color{Red}&space;\frac{x^{2}}{1+x^{5}}}={\color{Red}&space;\frac{{\color{Green}&space;{\color{DarkGreen}&space;a_{1}}{\color{DarkGreen}&space;.f_{1}\left&space;(&space;x{\color{Red}&space;}&space;\right&space;){\color{Red}&space;+{\color{Blue}&space;\left&space;(&space;a_{2}x+b_{2}&space;\right&space;)f_{2}\left&space;(&space;x&space;\right&space;){\color{Red}&space;+{\color{Magenta}&space;\left&space;(&space;a_{3}x+b_{3}&space;\right&space;)f_{3}\left&space;(&space;x&space;\right&space;)}}}}}}}{{{\color{DarkGreen}&space;\color{Green}&space;{\color{DarkGreen}&space;\left&space;(x+1&space;\right&space;){\color{Blue}&space;\left&space;(&space;x^{2}&space;-2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1\right&space;){\color{Magenta}&space;\left&space;(&space;x^{2}&space;-2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)x+1\right&space;)}}}}}}}{\color{Red}&space;----\left&space;(&space;18&space;\right&space;)}$

$\dpi{150}&space;{\color{DarkGreen}&space;a_{1}.f_{1}\left&space;(&space;x&space;\right&space;){\color{Red}&space;+{\color{Blue}&space;\left&space;(&space;a_{2}x+b_{2}&space;\right&space;)f_{2}\left&space;(&space;x&space;\right&space;){\color{Red}&space;+{\color{Magenta}&space;\left&space;(&space;a_{3}x+b_{3}&space;\right&space;)f_{3}\left&space;(&space;x&space;\right&space;)}}}}}={\color{Red}&space;x^{2}----\left&space;(&space;19&space;\right&space;)}$

$\dpi{150}&space;{\color{Red}&space;x=-1}\rightarrow&space;{\color{DarkGreen}&space;a_{1}.f_{1}\left&space;(&space;-1&space;\right&space;)}{\color{Red}&space;+}{\color{Blue}&space;\left&space;(&space;-a_{2}+b_{2}&space;\right&space;)f_{2}\left&space;(&space;-1&space;\right&space;)}{\color{Red}&space;+}{\color{Magenta}&space;\left&space;(&space;-a_{3}+b_{3}&space;\right&space;)f_{3}\left&space;(&space;-1&space;\right&space;)}={\color{Red}&space;1}$

$\dpi{150}&space;{\color{DarkGreen}&space;a_{1}.f_{1}\left&space;(&space;-1&space;\right&space;)={\color{Red}1&space;}}\left&space;(&space;because{\color{Blue}&space;\left&space;(&space;f_{2}\left&space;(&space;-1&space;\right&space;)=0&space;\right&space;)}and{\color{Magenta}&space;\left&space;(&space;f_{3}\left&space;(&space;-1&space;\right&space;)=0&space;\right&space;)}&space;\right&space;)$

$\dpi{150}&space;a_{1}\left&space;(&space;1+2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)+1&space;\right&space;)\left&space;(&space;1+2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)+1&space;\right&space;)=1$

$\dpi{150}&space;a_{1}\left&space;(&space;2+2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)\left&space;(&space;2+2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)&space;\right&space;)=1$

$\dpi{150}&space;a_{1}\left&space;(&space;4\left&space;(&space;\frac{1+cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}{2}&space;\right&space;)&space;\right&space;)\left&space;(&space;4\left&space;(&space;\frac{1+cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)}{2}&space;\right&space;)&space;\right&space;)=1$

$\dpi{150}&space;16cos^{2}\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)cos^{2}\left&space;(&space;\frac{3\pi&space;}{10}&space;\right&space;)a_{1}=1$

$\dpi{150}&space;{\color{DarkGreen}&space;a_{1}}={\color{DarkGreen}&space;\frac{1}{16cos^{2}\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)cos^{2}\left&space;(&space;\frac{3\pi&space;}{10}&space;\right&space;)}}={\color{DarkGreen}&space;\frac{1}{16}sec^{2}\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)sec^{2}\left&space;(&space;\frac{3\pi&space;}{10}&space;\right&space;)}{\color{Red}&space;---\left&space;(&space;20&space;\right&space;)}$

ويمكن التعويض عن (x) بقيم أخرى في المعادلة (19) لأيجاد (a2),(b2),(a3),(b3)

وبعد أيجاد القيم يمكن حل التكامل عن طريق تجزئة الكسور ولصعوبة حل السؤال قد يوضع في قائمة الأسئلة التي ليس لها حل

هذا ولكم جزيل الشكر والتقدير

2. ## رد: بعض التكاملات التي لا يمكن ان تحل

تعديل المعادلة (17) لوجود خطأ في كتابتها :

$\dpi{150}&space;{\color{Red}&space;\frac{x^{2}}{1+x^{5}}}={\color{Red}&space;\frac{x^{2}}{x^{5}+1}}={\color{DarkGreen}&space;\frac{a_{1}}{\left&space;(&space;x+1&space;\right&space;)}}+{\color{Blue}&space;\frac{a_{2}x+b_{2}}{\left&space;(&space;x^{2}-2cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}}+{\color{Magenta}&space;\frac{a_{3}x+b_{3}}{\left&space;(&space;x^{2}-2cos\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}}{\color{Red}&space;--\left&space;(&space;17&space;\right&space;)}$

3. ## رد: بعض التكاملات التي لا يمكن ان تحل

لا علاقة لطول الحل بالسؤال قابل أو غير للحل
جد لي الناتج ، ثم جد المشتقة لتصل للدالة الاصل

4. ## The Following User Says Thank You to كامل موسى الناصري For This Useful Post:

(11-11-2014)

5. ## رد: بعض التكاملات التي لا يمكن ان تحل

تكملة حل سؤال (12) :
معلومات مفيدة :

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{Blue}&space;\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)=\sin&space;\left&space;(&space;\frac{3\pi&space;}{10}&space;\right&space;)=\frac{\sqrt{5}+1}{4}}&space;\\&space;{\color{Blue}&space;\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)=\sin&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)=\frac{\sqrt{5}-1}{4}}$

نجد ( a1 ) بطريقة أخرى كالتالي :

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{black}&space;\\&space;a_{1}\left&space;(&space;4\left&space;(&space;\frac{1+\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}{2}&space;\right&space;)&space;\right&space;)\left&space;(&space;4\left&space;(&space;\frac{1+\cos&space;\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)}{2}&space;\right&space;)&space;\right&space;)=1\rightarrow&space;a_{1}\left&space;(&space;4\left&space;(&space;\frac{1+\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}{2}&space;\right&space;)&space;\right&space;)\left&space;(&space;4\left&space;(&space;\frac{1-\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}{2}&space;\right&space;)&space;\right&space;)=1}&space;\\&space;\rightarrow&space;{\color{black}a_{1}&space;\left&space;(&space;4\left&space;(&space;\frac{1+\frac{\sqrt{5}+1}{4}}{2}&space;\right&space;)&space;\right&space;)&space;\left&space;(&space;4\left&space;(&space;\frac{1-\frac{\sqrt{5}-1}{4}}{2}&space;\right&space;)&space;\right&space;)=1\rightarrow&space;a_{1}\left&space;(&space;\frac{5+\sqrt{5}}{2}&space;\right&space;)\left&space;(&space;\frac{5-\sqrt{5}}{2}&space;\right&space;)=1}$

$\dpi{150}&space;\fn_cm&space;\\&space;\rightarrow&space;a_{1}\left&space;(&space;\frac{25-5}{4}&space;\right&space;)=1\rightarrow&space;5a_{1}=1\rightarrow&space;{\color{DarkGreen}&space;a_{1}=\frac{1}{5}}{\color{Red}\&space;\&space;---\left&space;(&space;21&space;\right&space;)}&space;\\&space;in&space;\&space;{\color{Red}&space;\left&space;(&space;19&space;\right&space;)}&space;\&space;\&space;Let&space;\&space;\&space;{\color{Red}&space;x=e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}}\rightarrow&space;{\color{DarkGreen}&space;a_{1}.f_{1}\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}{\color{Red}&space;+}{\color{Blue}&space;\left&space;(&space;a_{2}&space;.e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}+b_{2}\right&space;)f_{2}\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}{\color{Red}&space;+}{\color{Magenta}&space;\left&space;(&space;a_{3}&space;.e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}+b_{3}\right&space;)f_{3}\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}{\color{Red}&space;=e^{\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)i}}$

$\dpi{150}&space;\fn_cm&space;\\&space;\therefore&space;\&space;{\color{Blue}&space;\left&space;(&space;a_{2}.e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}+b_{2}&space;\right&space;)f2\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}{\color{Red}&space;=e^{\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)i}}&space;\&space;because&space;\&space;{\color{DarkGreen}&space;\left&space;(&space;f_{1}\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)=0\right&space;)}&space;\&space;and&space;\&space;{\color{Magenta}&space;\left&space;(&space;f_{3}&space;\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)\right&space;)=0}&space;\\&space;{\color{Blue}&space;f_{2}\left&space;(&space;x&space;\right&space;)}={\color{DarkGreen}&space;\left&space;(&space;x+1&space;\right&space;)}{\color{Magenta}&space;\left&space;(&space;x-e^{\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)\left&space;(&space;x-e^{\left&space;(&space;\frac{-3\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}{\color{Red}&space;\&space;---\left&space;(&space;22&space;\right&space;)}$

$\dpi{150}&space;\fn_cm&space;\\&space;\therefore&space;\&space;{\color{Blue}&space;f_{2}\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}={\color{DarkGreen}&space;\left&space;(e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}+&space;1&space;\right&space;)}{\color{Magenta}&space;\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}-&space;e^{\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)i}\right&space;)\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}-e^{\left&space;(&space;\frac{-3\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}$

$\dpi{150}&space;\fn_cm&space;{\color{Blue}&space;\left&space;(&space;a_{2}&space;.e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}+b_{2}\right&space;)}{\color{DarkGreen}&space;\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}+1&space;\right&space;)}{\color{Magenta}&space;\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}-e^{\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}-e^{\left&space;(&space;\frac{-3\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}{\color{Red}&space;=e^{\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)i}}$

$\dpi{150}&space;\fn_cm&space;\\&space;\left&space;[&space;{\color{Blue}&space;e^{\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)i}}{\color{DarkGreen}&space;.e^{\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)i}}{\color{Magenta}&space;.e^{\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)i}&space;.e^{\left&space;(&space;\frac{-\pi&space;}{5}&space;\right&space;)i}}.{\color{Blue}&space;\left&space;(&space;a_{2}.&space;e^{\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)i}+b_{2}.&space;e^{\left&space;(&space;\frac{-\pi&space;}{10}&space;\right&space;)i}&space;\right&space;)}{\color{DarkGreen}&space;\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)i}+&space;e^{\left&space;(&space;\frac{-\pi&space;}{10}&space;\right&space;)i}\right&space;)}{\color{Magenta}&space;\left&space;(&space;e^{\left&space;(&space;\frac{-\pi&space;}{5}&space;\right&space;)i}&space;-&space;e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)\left&space;(&space;e^{\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)i}&space;-&space;e^{\left&space;(&space;\frac{-2\pi&space;}{5}&space;\right&space;)i}\right&space;)=&space;e^{\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)i}}\right&space;]$

$\dpi{150}&space;\fn_cm&space;\therefore&space;{\color{Blue}&space;\left&space;(&space;a_{2}.e^{\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)i}+b_{2}&space;.e^{\left&space;(&space;\frac{-\pi&space;}{10}&space;\right&space;)i}\right&space;)}{\color{DarkGreen}&space;\left&space;(&space;e^{\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)i}+e^{\left&space;(&space;\frac{-\pi&space;}{10}&space;\right&space;)i}&space;\right&space;)}{\color{Magenta}&space;\left&space;(&space;e^{\left&space;(&space;\frac{-\pi&space;}{5}&space;\right&space;)i}-e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)\left&space;(&space;e^{\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)i}-e^{\left&space;(&space;\frac{-2\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)=1}$

6. ## رد: بعض التكاملات التي لا يمكن ان تحل

تكملة حل سؤال (12)

$\dpi{150}&space;\fn_cm&space;\\&space;\left&space;(a_{2}.e^{\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)i}&space;+b_{2}.e^{\left&space;(&space;\frac{-\pi&space;}{10}&space;\right&space;)i}&space;\right&space;)=\frac{1}{\left&space;(e^{\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)i}+e^{\left&space;(&space;\frac{-\pi&space;}{10}&space;\right&space;)i}&space;\right&space;)\left&space;(e^{\left&space;(&space;\frac{-\pi&space;}{5}&space;\right&space;)i}-e^{\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)\left&space;(e^{\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)i}-e^{\left&space;(&space;\frac{-2\pi&space;}{5}&space;\right&space;)i}&space;\right&space;)}&space;\\&space;\\&space;=\frac{1}{\left&space;(&space;2&space;\right&space;)\left&space;(&space;-2&space;\right&space;i)\left&space;(&space;2i&space;\right&space;)\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)\sin&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)\sin&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}$

$\dpi{150}&space;\fn_cm&space;\\&space;\therefore&space;\&space;{\color{Blue}&space;\left&space;(&space;a_{2}&space;\left&space;(&space;\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)+i\sin&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)&space;\right&space;)+b_{2}\left&space;(&space;\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)-i\sin&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)&space;\right&space;)\right&space;)}&space;=\frac{1}{8\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)\left&space;(\sin&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)\sin&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)}&space;\\&space;{\color{black}=\frac{1}{4\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)\left&space;(&space;\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)-\cos&space;\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)&space;\right&space;)}=\frac{1}{4\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)\left&space;(&space;\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)+\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)}&space;}$

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{Blue}&space;\therefore&space;\&space;a_{2}&space;\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)&space;+i.a_{2}\sin&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)+b_{2}\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)-i.b_{2}\sin&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)}=\frac{1}{4\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)\left&space;(&space;\frac{\sqrt{5}+1}{4}&space;+\frac{\sqrt{5}-1}{4}\right&space;)}&space;\\&space;=\frac{1}{4\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)\left&space;(&space;\frac{2\sqrt{5}}{4}&space;\right&space;)}=\frac{1\left&space;(&space;\sqrt{5}&space;\right&space;)}{2\sqrt{5}\left&space;(&space;\sqrt{5}&space;\right&space;)\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)}=\frac{\sqrt{5}}{10\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)}&space;\\$

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{Blue}&space;\therefore&space;\left&space;(a_{2}&space;+b_{2}&space;\right&space;)\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)+i\left&space;(&space;a_{2}&space;-b_{2}\right&space;)\sin&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)=\frac{\sqrt{5}}{10\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)}+\left&space;(0&space;\right&space;)i&space;\&space;....\left&space;(&space;23&space;\right&space;)}&space;\\&space;\rightarrow&space;{\color{Blue}&space;\left&space;(&space;a_{2}+b_{2}&space;\right&space;)\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)=\frac{\sqrt{5}}{10\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)}}&space;\&space;\&space;and&space;\&space;\&space;{\color{Blue}\left&space;(&space;a_{2}-b_{2}&space;\right&space;)\sin&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)=0}&space;\&space;\&space;because&space;\&space;\&space;{\color{Blue}&space;\left&space;(&space;a_{2}\in&space;R&space;\&space;,&space;\&space;b_{2}\in&space;R&space;\right&space;)}$

$\dpi{150}&space;\fn_cm&space;\\&space;\because&space;\&space;{\color{Blue}&space;\left&space;(&space;a_{2}&space;+b_{2}\right&space;)\cos&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)=\frac{\sqrt{5}}{10\cos\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)&space;}}\rightarrow&space;{\color{Blue}\left&space;(&space;a_{2}&space;+b_{2}\right&space;)=\frac{\sqrt{5}}{10\cos^{2}&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)}&space;\&space;....\left&space;(&space;24&space;\right&space;)}\\&space;\because&space;\&space;{\color{Blue}&space;\left&space;(&space;a_{2}-b_{2}&space;\right&space;)\sin&space;\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)=0&space;}\rightarrow&space;{\color{Blue}&space;\left&space;(&space;a_{2}&space;-b_{2}\right&space;)=0}\rightarrow&space;{\color{Blue}&space;a_{2}=b_{2}&space;\&space;\&space;....\left&space;(&space;25&space;\right&space;)}&space;\\&space;\therefore&space;\&space;{\color{Blue}&space;\left&space;(&space;b_{2}&space;+b_{2}\right&space;)=\frac{\sqrt{5}}{10\cos&space;^{2}\left&space;(&space;\frac{\pi&space;}{10}&space;\right&space;)}}=\frac{\sqrt{5}}{5\left&space;(&space;1+\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)}$

$\dpi{150}&space;\fn_cm&space;\\&space;\therefore&space;{\color{Blue}&space;2b_{2}}=\frac{\sqrt{5}}{5\left&space;(&space;1+\frac{1+\sqrt{5}}{4}&space;\right&space;)}=\frac{\sqrt{5}}{5\left&space;(&space;\frac{5+\sqrt{5}}{4}&space;\right&space;)}=\frac{4\sqrt{5}\left&space;(&space;5-\sqrt{5}&space;\right&space;)}{5\left&space;(&space;5+\sqrt{5}&space;\right&space;)\left&space;(&space;5-\sqrt{5}&space;\right&space;)}=\frac{20\sqrt{5}-20}{5\left&space;(&space;25-5&space;\right&space;)}{\color{Black}&space;=\frac{20\left&space;(&space;\sqrt{5}&space;-1\right&space;)}{5\left&space;(&space;20&space;\right&space;)}&space;}{\color{Blue}&space;=\frac{\left&space;(&space;\sqrt{5}-1&space;\right&space;)}{5}}&space;\\&space;\\&space;\therefore&space;\&space;\&space;{\color{Blue}&space;a_{2}=b_{2}=\frac{\left&space;(&space;\sqrt{5}&space;-1\right&space;)}{10}&space;\&space;\&space;....\left&space;(&space;26&space;\right&space;)}$

$\dpi{150}&space;\fn_cm&space;\\&space;in&space;\&space;{\color{Red}&space;\left&space;(&space;19&space;\right&space;)}\&space;\&space;Let&space;{\color{Red}&space;\&space;\&space;x=0}\rightarrow&space;{\color{DarkGreen}&space;a_{1}.f_{1}\left&space;(&space;0&space;\right&space;)}+{\color{Blue}\left&space;(&space;a_{2}\left&space;(&space;0&space;\right&space;)&space;+b_{2}\right&space;)f_{2}\left&space;(&space;0&space;\right&space;)}+{\color{Magenta}&space;\left&space;(&space;a_{3}&space;\left&space;(&space;0&space;\right&space;)+b_{3}\right&space;)f_{3}\left&space;(&space;0&space;\right&space;)}={\color{Red}&space;0^{2}}&space;\\&space;\therefore&space;\&space;{\color{DarkGreen}&space;a_{1}}+{\color{Blue}&space;b_{2}}+{\color{Magenta}&space;b_{3}}={\color{Red}&space;0}&space;\&space;\left&space;(&space;because&space;\&space;\&space;{\color{DarkGreen}&space;f_{1}\left&space;(&space;0&space;\right&space;)}={\color{Blue}&space;f_{2}\left&space;(&space;0&space;\right&space;)}&space;={\color{Magenta}&space;f_{3}\left&space;(&space;0&space;\right&space;)}={\color{Red}&space;1}\right&space;)$

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{Magenta}b_{_{3}}&space;}=-{\color{DarkGreen}&space;a_{1}}-{\color{Blue}&space;b_{2}}=\frac{-1}{5}-\frac{\sqrt{5}-1}{10}=\frac{-2-\sqrt{5}+1}{10}&space;\\&space;\therefore&space;\&space;\&space;{\color{Magenta}b_{3}&space;=\frac{-\sqrt{5}-1}{10}&space;\&space;.....\left&space;(&space;27&space;\right&space;)}&space;\\&space;in&space;\&space;{\color{Red}&space;\left&space;(&space;19&space;\right&space;)}&space;\&space;\&space;Let&space;\&space;\&space;{\color{Red}&space;x=1}\rightarrow&space;{\color{DarkGreen}&space;a_{1}.f_{1}\left&space;(&space;1&space;\right&space;)}+{\color{Blue}&space;\left&space;(&space;a_{2}\left&space;(&space;1&space;\right&space;)+b_{2}&space;\right&space;)f_{2}\left&space;(&space;1&space;\right&space;)}+{\color{Magenta}&space;\left&space;(&space;a_{3}\left&space;(&space;1&space;\right&space;)+b_{3}&space;\right&space;)f_{3}\left&space;(&space;1&space;\right&space;)}={\color{Red}&space;1^{2}}$

7. ## رد: بعض التكاملات التي لا يمكن ان تحل

تكملة حل سؤال ( 12):

$\dpi{150}&space;\fn_cm&space;\\&space;\therefore&space;\&space;{\color{DarkGreen}&space;\frac{1}{5}.\left&space;(&space;2-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)\left&space;(&space;2-2\cos&space;\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)&space;\right&space;)}+{\color{Blue}&space;\left&space;(&space;\frac{\left&space;(&space;\sqrt{5}-1&space;\right&space;)}{10}&space;+\frac{\left&space;(&space;\sqrt{5}-1&space;\right&space;)}{10}\right&space;)\left&space;(2&space;\right&space;)\left&space;(&space;2-2\cos&space;\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)&space;\right&space;)}{\color{Magenta}&space;+\left&space;(a_{3}&space;+\frac{\left&space;(&space;-\sqrt{5}&space;-1\right&space;)}{10}&space;\right&space;)\left&space;(2&space;\right&space;)\left&space;(&space;2-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)=1}$

$\dpi{150}&space;\\&space;\frac{1}{5}\left&space;(&space;4-4\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)\left&space;(&space;1+\cos&space;\left&space;(\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)+\frac{1}{5}\left&space;(&space;4\sqrt{5}&space;-4\right&space;)\left&space;(&space;1+\cos&space;\left&space;(\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)+\left&space;(&space;a_{3}+\frac{\left&space;(&space;-\sqrt{5}-1&space;\right&space;)}{10}&space;\right&space;)\left&space;(&space;4-4\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;){\color{black}&space;=1}&space;\\&space;\frac{1}{5}\left&space;(&space;1+\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)\left&space;(&space;4\sqrt{5}-4\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)+\left&space;(a_{3}+\frac{\left&space;(&space;-\sqrt{5}&space;-1\right&space;)}{10}&space;\right&space;)\left&space;(&space;4-4\left&space;(&space;\frac{\sqrt{5}+1}{4}&space;\right&space;)&space;\right&space;){\color{black}&space;=1}$

$\dpi{150}&space;\\&space;\left&space;[\\&space;\frac{1}{5}\left&space;(&space;1+\frac{\sqrt{5}-1}{4}&space;\right&space;)\left&space;(&space;4\sqrt{5}-\frac{4\left&space;(&space;\sqrt{5}+1&space;\right&space;)}{4}&space;\right&space;)+\left&space;(&space;a_{3}-\frac{\sqrt{5}+1}{10}&space;\right&space;)\left&space;(&space;3-\sqrt{5}&space;\right&space;)=1&space;\right&space;]{\color{Red}&space;\left&space;(&space;20&space;\right&space;)}&space;\\&space;\\&space;\rightarrow&space;\left&space;(&space;4+\sqrt{5}-1&space;\right&space;)\left&space;(&space;4\sqrt{5}&space;-\sqrt{5}-1\right&space;)+\left&space;(&space;20a_{3}&space;-2\sqrt{5}-2\right&space;)\left&space;(&space;3-\sqrt{5}&space;\right&space;)=20&space;\\&space;\rightarrow&space;\left&space;[&space;\left&space;(&space;3+\sqrt{5}&space;\right&space;)\left&space;(&space;3\sqrt{5}&space;-1\right&space;)+\left&space;(&space;20a_{3}&space;-2\sqrt{5}-2\right&space;)\left&space;(&space;3-\sqrt{5}&space;\right&space;)=20&space;\right&space;]{\color{Red}&space;\left&space;(&space;3+\sqrt{5}&space;\right&space;)}&space;\\&space;\rightarrow&space;\left&space;(&space;8\sqrt{5}+12&space;\right&space;)\left&space;(&space;\sqrt{5}+3&space;\right&space;)+4\left&space;(&space;20a_{3}-2\sqrt{5}&space;-2\right&space;)=60+20\sqrt{5}$

$\dpi{150}&space;\\&space;\rightarrow&space;\left&space;(&space;76+36\sqrt{5}&space;\right&space;)+80a_{3}-8\sqrt{5}-8=60+20\sqrt{5}&space;\\&space;\\&space;\rightarrow&space;80a_{3}+68+28\sqrt{5}=60+20\sqrt{5}\rightarrow&space;80a_{3}=60+20\sqrt{5}-68-28\sqrt{5}&space;\\&space;\rightarrow&space;80a_{3}=-8\sqrt{5}-8&space;\\&space;\rightarrow&space;{\color{Magenta}&space;a_{3}=b_{3}=\frac{\left&space;(&space;-\sqrt{5}&space;-1\right&space;)}{10}....\left&space;(&space;28\right&space;)}$

نعوض عن القيم a1,a2,a3,b1,b2,b3 في المعادلة رقم (17) (بعد التعديل)

$\dpi{150}&space;\&space;\therefore&space;{\color{Red}\frac{x^{2}}{1+x^{5}}&space;}={\color{DarkGreen}&space;\frac{\left&space;(&space;\frac{1}{5}&space;\right&space;)}{\left&space;(&space;x+1&space;\right&space;)}}+{\color{Blue}&space;\frac{\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)x+\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)}{\left&space;(&space;x^{2}&space;-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1\right&space;)}}+{\color{Magenta}&space;\frac{\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)x+\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)}{\left&space;(x^{2}-2\cos&space;\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}&space;}&space;\&space;\&space;{\color{Red}&space;....\left&space;(&space;29&space;\right&space;)}$

التحقق من صحة المتطابقة رقم (29) ( مختصر مفيد):

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{Blue}&space;R.H.S=}{\color{DarkGreen}&space;\frac{\left&space;(&space;\frac{1}{5}&space;\right&space;)}{\left&space;(&space;x+1&space;\right&space;)}}+{\color{Blue}&space;\frac{\left&space;(&space;\frac{\sqrt{5}-1}{10}\right&space;)x+\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)}{\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}}+{\color{Magenta}&space;\frac{\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)x+\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)}{\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}&space;}$

$\dpi{150}&space;\fn_cm&space;\\&space;=&space;{\color{DarkGreen}&space;\frac{\left&space;(&space;\frac{1}{5}&space;\right&space;)}{\left&space;(&space;x+1&space;\right&space;)}}+{\color{Blue}&space;\frac{\left&space;(&space;\frac{\sqrt{5}-1}{10}\right&space;)x+\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)}{\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}}+{\color{Magenta}&space;\frac{\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)x+\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)}{\left&space;(&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}&space;}&space;\\&space;\\$

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{Black}R.H.S=&space;\frac{\left&space;(&space;\frac{1}{5}&space;\right&space;)\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)\left&space;(&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)+\left&space;(&space;\frac{\sqrt{5}-1}{10}\right&space;)\left&space;(&space;x+1&space;\right&space;)^{2}\left&space;(&space;x^{2}&space;+2\cos\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;x+1\right&space;)-\left&space;(&space;\frac{\sqrt{5}+1}{10}&space;\right&space;)\left&space;(&space;x+1&space;\right&space;)^{2}\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}{\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)\left&space;(&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}}...(30)$

8. ## رد: بعض التكاملات التي لا يمكن ان تحل

تكملة حل سؤال (12) :

$\dpi{150}&space;\fn_cm&space;\\&space;\because&space;\&space;{\color{Blue}&space;\left&space;(&space;x^{2}&space;-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1\right&space;)\left&space;(&space;x^{2}&space;+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1\right&space;)}&space;\\&space;\\&space;=\left&space;(&space;x^{4}+\left&space;(&space;2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;-2\cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)x^{3}&space;+\left&space;(&space;1-4\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;+1\right&space;)x^{2}+\left&space;(&space;2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)-2\cos\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)x+1\right&space;)$

$\dpi{150}&space;\fn_cm&space;\\&space;=\left&space;(&space;x^{4}+\left&space;(&space;\frac{\sqrt{5}-1}{2}-\frac{\sqrt{5}+1}{2}&space;\right&space;)x^{3}+\left&space;(&space;2-\left&space;(&space;\sqrt{5}+1&space;\right&space;)\left&space;(&space;\frac{\sqrt{5}-1}{4}&space;\right&space;)&space;\right&space;)&space;x^{2}+\left&space;(&space;\frac{\sqrt{5}-1}{2}-\frac{\sqrt{5}+1}{2}\right&space;)x+1\right&space;)&space;\\&space;\\&space;={\color{Blue}&space;\left&space;(&space;x^{4}&space;-x^{3}+x^{2}-x+1\right&space;)}$

$\dpi{150}&space;\fn_cm&space;\\&space;\because&space;\&space;{\color{Red}&space;\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;x+1&space;\right&space;)^{2}\left&space;(&space;x^{2}&space;+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1\right&space;)-\left&space;(&space;\frac{\sqrt{5}+1}{10}&space;\right&space;)\left&space;(&space;x+1&space;\right&space;)^{2}\left&space;(&space;x^{2}&space;-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1\right&space;)}&space;\\&space;\\&space;=\left&space;(&space;x+1&space;\right&space;)^{2}\left&space;[&space;\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)-\left&space;(&space;\frac{\sqrt{5}+1}{10}&space;\right&space;)\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)&space;\right&space;]$

$\dpi{150}&space;\fn_cm&space;\\&space;=\left&space;(&space;x+1&space;\right&space;)^{2}\left&space;[&space;\left&space;(&space;\frac{\sqrt{5}-1}{10}-\frac{\sqrt{5}+1}{10}&space;\right&space;)x^{2}+\left&space;(&space;\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;\frac{\sqrt{5}-1}{2}&space;\right&space;)+\left&space;(&space;\frac{\sqrt{5}+1}{10}&space;\right&space;)\left&space;(&space;\frac{\sqrt{5}+1}{2}&space;\right&space;)&space;\right&space;)&space;x+\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;-\frac{\sqrt{5}+1}{10}\right&space;)\right&space;]&space;\\&space;\\&space;=\left&space;(&space;x+1&space;\right&space;)^{2}\left&space;[&space;\left&space;(\frac{-1}{5}&space;\right&space;)x^{2}+\left&space;(&space;\frac{6-2\sqrt{5}}{20}+\frac{6+2\sqrt{5}}{20}&space;\right&space;)x-\frac{1}{5}&space;\right&space;]&space;\\&space;\\&space;=\left&space;(&space;x^{2}+2x+1&space;\right&space;)\left&space;\left&space;[&space;\left&space;(&space;\frac{-1}{5}&space;\right&space;)x^{2}+\left&space;(&space;\frac{3}{5}&space;\right&space;)x-\frac{1}{5}&space;\right&space;]$

$\dpi{150}&space;\fn_cm&space;\\&space;=\left&space;[&space;\left&space;(&space;\frac{-1}{5}&space;\right&space;)x^{4}&space;+\left&space;(&space;\frac{3}{5}&space;-\frac{2}{5}\right&space;)x^{3}+\left&space;(&space;\frac{-1}{5}+\frac{6}{5}&space;-\frac{1}{5}\right&space;)x^{2}+\left&space;(&space;\frac{3}{5}&space;-\frac{2}{5}\right&space;)x-\frac{1}{5}\right&space;]&space;\\&space;\\&space;={\color{Red}&space;\left&space;[&space;\left&space;(&space;\frac{-1}{5}&space;\right&space;)&space;x^{4}+\left&space;(&space;\frac{1}{5}&space;\right&space;)x^{3}+\left&space;(&space;\frac{4}{5}&space;\right&space;)x^{2}+\left&space;(&space;\frac{1}{5}&space;\right&space;)x-\frac{1}{5}\right&space;]}$

$\dpi{150}&space;\fn_cm&space;\\&space;\therefore&space;\&space;{\color{Blue}&space;R.H.S}&space;=\frac{\left&space;(&space;\frac{1}{5}&space;\right&space;){\color{Blue}&space;\left&space;(&space;x^{4}-x^{3}+x^{2}-x+1&space;\right&space;)}+{\color{Red}&space;\left&space;[&space;\left&space;(&space;\frac{-1}{5}&space;\right&space;)x^{4}+\left&space;(&space;\frac{1}{5}&space;\right&space;)x^{3}+\left&space;(&space;\frac{4}{5}&space;\right&space;)x^{2}&space;+\left&space;(&space;\frac{1}{5}&space;\right&space;)x-\frac{1}{5}\right&space;]}}{\left&space;(&space;x+1&space;\right&space;){\color{Blue}&space;\left&space;(&space;x^{4}-x^{3}+x^{2}-x+1&space;\right&space;)}}&space;\\&space;\\&space;=\frac{\left&space;(&space;\frac{1}{5}-\frac{1}{5}&space;\right&space;)x^{4}+\left&space;(&space;\frac{-1}{5}+\frac{1}{5}&space;\right&space;)x^{3}+\left&space;(&space;\frac{1}{5}+\frac{4}{5}&space;\right&space;)x^{2}+\left&space;(&space;\frac{-1}{5}&space;+\frac{1}{5}\right&space;)x+\left&space;(&space;\frac{1}{5}&space;-\frac{1}{5}\right&space;)}{\left&space;(&space;x^{5}&space;+1\right&space;)}$

$\dpi{150}&space;\fn_cm&space;={\color{Red}&space;\frac{x^{2}}{1+x^{5}}}={\color{Blue}&space;L.H.S}$

$\dpi{150}&space;\fn_cm&space;\\&space;\because&space;{\color{Red}&space;\frac{x^{2}}{1+x^{5}}}={\color{DarkGreen}&space;\frac{\left&space;(\frac{1}{5}&space;\right&space;)}{\left&space;(&space;x+1&space;\right&space;)}}+{\color{Blue}&space;\frac{\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)x+\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)}{\left&space;(x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;x+1\right&space;)}}+{\color{Magenta}&space;\frac{\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)x+\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)}{\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{3\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}}$

$\dpi{150}&space;\fn_cm&space;\therefore&space;\&space;\&space;{\color{Red}&space;\int&space;\frac{x^{2}}{1+x^{5}}dx}={\color{DarkGreen}\left&space;(&space;\frac{1}{5}&space;\right&space;)&space;\int&space;\frac{1}{\left&space;(&space;x+1&space;\right&space;)}dx}+{\color{Blue}&space;\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\int&space;\frac{\left&space;(&space;x+1&space;\right&space;)}{\left&space;(&space;x^{2}&space;-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1\right&space;)}dx}{\color{Magenta}&space;+\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)\int&space;\frac{\left&space;(&space;x+1&space;\right&space;)}{\left&space;(x^{2}&space;+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}dx \ ...(31)}$

$\dpi{150}&space;\fn_cm&space;{\color{DarkGreen}&space;\left&space;(&space;\frac{1}{5}&space;\right&space;)\int&space;\frac{1}{\left&space;(&space;x+1&space;\right&space;)}dx}={\color{DarkGreen}&space;\frac{1}{5}\ln&space;\left&space;|&space;x+1&space;\right&space;|+c&space;\&space;\&space;....\left&space;(&space;32&space;\right&space;)&space;}$

9. ## رد: بعض التكاملات التي لا يمكن ان تحل

تكملة حل سؤال (12):

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{Blue}\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)&space;\int&space;\frac{\left&space;(&space;x+1&space;\right&space;)}{\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}dx}=\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\int&space;\left&space;(&space;\frac{\left&space;(&space;x-\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)}{\left&space;(&space;x^{2}&space;-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1\right&space;)}+\frac{\left&space;(&space;1+\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)}{\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}&space;\right&space;)dx$

$\dpi{150}&space;\fn_cm&space;\\&space;=\left&space;(&space;\frac{\sqrt{5}-1}{20}&space;\right&space;)&space;\int&space;\frac{\left&space;(&space;2x-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)}{\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}dx+\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;1+\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)\int&space;\frac{1}{\left&space;(&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x&space;+\cos&space;^{2}\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)\right&space;)+\sin&space;^{2}\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}dx$

$\dpi{150}&space;\\&space;=\left&space;(&space;\frac{\sqrt{5}-1}{20}&space;\right&space;)\ln&space;\left&space;|x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;|+\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;1+\frac{\sqrt{5}+1}{4}&space;\right&space;)\int&space;\frac{1}{\left&space;(&space;x-\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)^{2}+\sin&space;^{2}\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}dx$

$\dpi{150}&space;\fn_cm&space;\\&space;=\left&space;(&space;\frac{\sqrt{5}-1}{20}&space;\right&space;)\ln&space;\left&space;|&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;x+1\right&space;|+\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;\frac{\sqrt{5}+5}{4}&space;\right&space;)\left&space;(&space;\frac{1}{\sin&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}&space;\right&space;)\int&space;\frac{\left&space;(&space;\frac{1}{\sin&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}&space;\right&space;)}{\left&space;[&space;\left&space;(&space;\frac{x}{\sin&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}&space;-\cot&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)\right&space;)&space;^{2}+1\right&space;]}dx$

$\dpi{150}&space;\fn_cm&space;\\&space;\therefore&space;\&space;\&space;{\color{Blue}&space;\left&space;(&space;\frac{\sqrt{5}-1}{10}&space;\right&space;)\int&space;\frac{\left&space;(x+1&space;\right&space;)}{\left&space;(x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}dx=\left&space;(&space;\frac{\sqrt{5}-1}{20}&space;\right&space;)\ln&space;\left&space;|&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;|+\left&space;(\frac{\sqrt{5}\csc&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}{10}&space;\right&space;)\tan^{-1}\left&space;(&space;\csc&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x-\cot&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)+c \ ...(33)}$

$\dpi{150}&space;\fn_cm&space;\\&space;{\color{Magenta}&space;\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)\int\frac{\left&space;(&space;x+1\right&space;)}{\left&space;(&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1\right&space;)}dx}=\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)\int&space;\left&space;(&space;\frac{\left&space;(&space;x+\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)\right&space;)}{\left&space;(&space;x^{2}&space;+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1\right&space;)}&space;+\frac{\left&space;(1-\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)}{\left&space;(x^{2}&space;+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}\right&space;)dx$

$\dpi{150}&space;\fn_cm&space;=\left&space;(&space;\frac{-\sqrt{5}-1}{20}&space;\right&space;)\int&space;\frac{\left&space;(&space;2x+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)}{\left&space;(&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;x+1\right&space;)}dx+\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;1-\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)\int&space;\frac{1}{\left&space;(&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+\cos^{2}\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)+\sin&space;^{2}\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}dx$

$\dpi{150}&space;\fn_cm&space;=\left&space;(&space;\frac{-\sqrt{5}-1}{20}&space;\right&space;)\ln&space;\left&space;|&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;x+1\right&space;|+\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;1-\frac{\left&space;(&space;\sqrt{5}&space;-1\right&space;)}{4}&space;\right&space;)\int&space;\frac{1}{\left&space;(&space;x+\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)^{2}+\sin&space;^{2}\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}dx$

$\dpi{150}&space;\fn_cm&space;=\left&space;(&space;\frac{-\sqrt{5}-1}{20}&space;\right&space;)\ln&space;\left&space;|&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;x+1\right&space;|+\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)\left&space;(&space;\frac{-\sqrt{5}+5}{4}&space;\right&space;)\left&space;(&space;\frac{1}{\sin&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}&space;\right&space;)\int&space;\frac{\left&space;(&space;\frac{1}{\sin&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}&space;\right&space;)}{\left&space;(&space;\frac{x}{\sin\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}&space;+\cot&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)\right&space;)^{2}+1}dx$

$\dpi{150}&space;\fn_cm&space;\therefore&space;{\color{Magenta}&space;\&space;\&space;\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)\int&space;\frac{\left&space;(&space;x+1&space;\right&space;)}{\left&space;(&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;)}dx&space;=&space;-\left&space;(&space;\frac{\sqrt{5}+1}{20}&space;\right&space;)\ln&space;\left&space;|&space;x^{2}+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;x+1\right&space;|-\left&space;(&space;\frac{\sqrt{5}\csc&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}{10}&space;\right&space;)\tan^{-1}\left&space;(&space;\csc&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+\cot&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)+c&space;\&space;...\left&space;(&space;34&space;\right&space;)}$

10. ## رد: بعض التكاملات التي لا يمكن ان تحل

تكملة حل سؤال (12) :

$\dpi{150}&space;\fn_cm&space;\\&space;\therefore&space;\&space;\&space;{\color{Red}&space;\int&space;\frac{x^{2}}{1+x^{5}}dx}{\color{DarkGreen}&space;=}{\color{DarkGreen}\left&space;(&space;\frac{1}{5}&space;\right&space;)&space;\int&space;\frac{1}{\left&space;(&space;x+1&space;\right&space;)}dx}{\color{Blue}&space;+}{\color{Blue}&space;\left&space;(&space;\frac{\sqrt{5}-1}{10&space;}&space;\right&space;)\int&space;\frac{\left&space;(&space;x+1&space;\right&space;)}{\left&space;(&space;x^{2}&space;-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1\right&space;)}dx}{\color{Magenta}&space;{\color{Blue}&space;{\color{Magenta}&space;+}}}{\color{Magenta}&space;\left&space;(&space;\frac{-\sqrt{5}-1}{10}&space;\right&space;)\int&space;\frac{\left&space;(&space;x+1&space;\right&space;)}{\left&space;(&space;x^{2}&space;+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1\right&space;)}dx}$

$\dpi{900}&space;\fn_cm&space;{\color{Red}&space;=\frac{1}{5}\ln&space;\left&space;|&space;x+1&space;\right&space;|+\left&space;(&space;\frac{\sqrt{5}-1}{20}&space;\right&space;)\ln&space;\left&space;|&space;x^{2}-2\cos&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x+1&space;\right&space;|+\left&space;(&space;\frac{\sqrt{5}\csc&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)}{10}&space;\right&space;)\tan^{-1}\left&space;(&space;\csc&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)x-\cot&space;\left&space;(&space;\frac{\pi&space;}{5}&space;\right&space;)&space;\right&space;)-\left&space;(&space;\frac{\sqrt{5}+1}{20}&space;\right&space;)\ln&space;\left&space;|&space;x^{2}&space;+2\cos&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+1\right&space;|-\left&space;(&space;\frac{\sqrt{5}\csc&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)}{10}&space;\right&space;)\tan^{-1}\left&space;(&space;\csc&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)x+\cot&space;\left&space;(&space;\frac{2\pi&space;}{5}&space;\right&space;)&space;\right&space;)+c}$

11. ## رد: بعض التكاملات التي لا يمكن ان تحل

السلام عليكم استاذي العزيز ..
بالنسبة لتكاملات e^x^2 بحثت في النت و وجدت انها تسمى تكاملات غاوس وهذا رابط الشرح من ويكيبيديا
https://ar.wikipedia.org/wiki/%D8%AA...88%D8%B3%D9%8A

13. ## The Following User Says Thank You to حاجم الربيعي For This Useful Post:

(05-24-2017)

صفحة 2 من 2 الأولى 12

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