بعض التكاملات التي لا يمكن ان تحل

رد: بعض التكاملات التي لا يمكن ان تحل

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

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

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

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

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

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

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

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

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

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

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

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

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

$a_{1}\left ( 1+2cos\left ( \frac{\pi }{5} \right )+1 \right )\left ( 1+2cos\left ( \frac{3\pi }{5} \right )+1 \right )=1$

$a_{1}\left ( 2+2cos\left ( \frac{\pi }{5} \right ) \right )\left ( 2+2cos\left ( \frac{3\pi }{5} \right ) \right )=1$

$a_{1}\left ( 4\left ( \frac{1+cos\left ( \frac{\pi }{5} \right )}{2} \right ) \right )\left ( 4\left ( \frac{1+cos\left ( \frac{3\pi }{5} \right )}{2} \right ) \right )=1$

$16cos^{2}\left ( \frac{\pi }{10} \right )cos^{2}\left ( \frac{3\pi }{10} \right )a_{1}=1$

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

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

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