سؤال في التكامل

الملف الشخصي · 12-22-2014, 11:31 PM

جد ناتج التكامل التالي :

^{^{^{^{^{^{^{_{_{حيث k ثابت}}}}}}}}} ${\color{Blue} \int \frac{\left (x^{2}+4 \right )^{5}x^{k+7}}{\left ( x^{2} -4\right )^{k+13}}dx \ \ : \ k\in N}$

الملف الشخصي · 12-30-2014, 02:15 AM

جد ناتج التكامل التالي :

^{^{^{^{^{^{^{حيث k ثابت}}}}}}} ${\color{Blue} \int \frac{\left (x^{2}+4 \right )^{5}x^{k+7}}{\left ( x^{2} -4\right )^{k+13}}dx \ \ : \ k\in N}$

$\\ Let \ \ {\color{Blue} x=}2{\color{Blue} \left ( {\color{Blue} \sec \left ( y \right )+\tan \left ( y \right )} \right )}\rightarrow {\color{Blue} dx=}2{\color{Blue} \left ( \sec \left ( y \right )\tan \left ( y \right ) +\sec ^{2}\left ( y \right )\right )dy}=2{\color{Blue} \sec \left ( y \right )\left ( \tan \left ( y \right )+\sec \left ( y \right ) \right )dy} \\ \\ {\color{Blue} I=\int \frac{\left (x^{2}+4 \right )^{5}x^{k+7}}{\left (x^{2}-4 \right )^{k+13}}dx}$

$\\ =\int \frac {2\left [ \left ( 2\left ( \sec \left ( y \right )+\tan \left ( y \right ) \right ) \right )^{2}+4\right ]^{5}\left [ 2\left ( \sec \left ( y \right )+\tan \left ( y \right ) \right ) \right ]^{k+7}\sec \left ( y \right )\left ( \sec \left ( y\left \right ) +\tan \left ( y \right )\right )}{\left [ \left ( 2\left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )\right ) ^{2}-4\right ]^{k+13}}dy$

$\\ =\int \frac{2^{k+18}{\color{Blue} \left [ \left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )^{2} +1\right ]^{5}}\left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )^{k+7}\sec \left ( y \right )\left ( \sec \left ( y \right ) +\tan \left ( y \right )\right )}{2^{2k+26}{\color{Red} \left [ \left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )^{2} -1\right ]^{k+13}}}dy$

$\\ =\frac{1}{2^{k+8}}\int \frac{{\color{Blue} \left [\sec ^{2} \left ( y \right )+2\sec \left ( y \right )\tan \left ( y \right ) +\tan ^{2}\left ( y \right )+1\right ]^{5}}\left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )^{k+8}\sec \left ( y \right )}{{\color{Red} \left [ \sec ^{2}\left ( y \right )+2\sec \left ( y \right )\tan \left ( y \right )+\tan ^{2}\left ( y \right )-1 \right ]^{k+13}}}dy$

$\\ =\frac{1}{2^{k+8}}\int \frac{{\color{Blue} \left [ 2\sec ^{2}\left ( y \right )+2\sec \left ( y \right )\tan \left ( y \right ) \right ]^{5}}\left ( \sec \left ( y \right ) +\tan \left ( y \right )\right )^{k+8}\sec \left ( y \right )}{{\color{Red}\left [ 2\sec \left ( y \right )\tan \left ( y \right )+2\tan ^{2} \left ( y \right )\right ] ^{k+13}}}dy$

$\\ =\frac{1}{2^{k+8}}\int \frac{{\color{Blue}\left [ 2\sec \left ( y \right ) \left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )\right ]^{5} }\left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )^{k+8}\sec \left ( y \right )}{{\color{Red}\left [ 2\tan \left ( y \right )\left ( \sec \left ( y \right ) +\tan \left ( y \right )\right ) \right ]^{k+13} }}dy$

$\\ =\frac{1}{2^{k+8}}\int \frac{{\color{Blue}2^{5}\sec ^{5}\left ( y \right ) \left ( \sec \left ( y \right ) +\tan \left ( y \right )\right )^{5}}\left ( \sec \left ( y \right ) +\tan \left ( y \right )\right )^{k+8}\sec \left ( y \right )}{{\color{Red}2^{k+13} \tan ^{k+13}\left ( y \right )\left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )^{k+13}}}dy \\ \\ =\frac{1}{2^{2k+16}}\int \sec ^{6}\left ( y \right )\left ( \tan \left ( y \right ) \right )^{-k-13}dy=\frac{1}{2^{2k+16}}\int \sec ^{2}\left ( y \right )\left ( 1+\tan ^{2}\left ( y \right ) \right )^{2}\left ( \tan \left ( y \right ) \right )^{-k-13}dy$

$\\ =\frac{1}{2^{2k+16}}\int \sec ^{2}\left ( y \right )\left ( 1+2\tan ^{2}\left ( y \right )+\tan ^{4}\left ( y \right ) \right )\left ( \tan \left ( y \right ) \right )^{-k-13}dy \\ \\ =\frac{1}{2^{2k+16}}\int \left (\sec ^{2}\left ( y \right ) \left ( \tan \left ( y \right ) \right )^{-k-13} +2\sec ^{2}\left ( y \right )\left ( \tan \left ( y \right ) \right )^{-k-11}+\sec ^{2}\left ( y \right )\left ( \tan \left ( y \right ) \right )^{-k-9}\right )dy$

$\\ =\frac{1}{2^{2k+16}}\left ( \frac{\left ( \tan \left ( y \right ) \right )^{-k-12}}{-k-12}+\frac{2\left ( \tan \left ( y \right ) \right )^{-k-10}}{-k-10}+\frac{\left ( \tan \left ( y \right ) \right )^{-k-8}}{-k-8} \right )+c \\ \\ {\color{Blue} =\frac{-1}{2^{2k+16}}\left ( \frac{1}{\left ( k+12 \right )\left ( \tan \left ( y \right ) \right )^{k+12}}+\frac{2}{\left ( k+10 \right )\left ( \tan \left ( y \right ) \right )^{k+10}} +\frac{1}{\left ( k+8 \right )\left ( \tan \left ( y \right ) \right )^{k+8}}\right )+c}$

الملف الشخصي · 12-30-2014, 02:19 AM

تكملة الحل :

$\\ \because \ \ {\color{Blue} x=2\left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )}\rightarrow {\color{Blue} \left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )=\frac{x}{2} \ ...\left ( 1 \right )} \\ \\ \because \ \ \sec ^{2}\left ( y \right )-\tan ^{2}\left ( y \right )=1\rightarrow \left ( \sec \left ( y \right )+\tan \left ( y \right ) \right )\left ( \sec \left ( y \right )-\tan \left ( y \right ) \right )=1$

$\\ \therefore \ \ \frac{x}{2}\left ( \sec \left ( y \right )-\tan \left ( y \right ) \right )=1\rightarrow {\color{Red} \left ( \sec \left ( y \right ) -\tan \left ( y \right )\right )=\frac{2}{x}\ ...\left ( 2 \right )}$

وبطرح المعادلة (2) من المعادلة (1) ينتج أن :

$\\ 2\tan \left ( y \right ) =\frac{x}{2}-\frac{2}{x}\rightarrow {\color{Blue} \tan \left ( y \right )=\frac{x}{4}-\frac{1}{x}} \\ \\ \therefore \ {\color{Blue} I=\int \frac{\left ( x^{2}+4 \right )^{5}x^{k+7}}{\left ( x^{2}-4 \right )^{k+13}}dx} \\ \\ =\frac{-1}{2^{2k+16}}\left ( \frac{1}{\left ( k+12 \right )\left ( \tan \left ( y \right ) \right )^{k+12}} +\frac{2}{\left ( k+10 \right )\left ( \tan \left ( y \right ) \right )^{k+10}}+\frac{1}{\left ( k +8\right )\left ( \tan \left ( y \right ) \right )^{k+8}}\right ){\color{Black}+c }$

$\\ =\frac{-1}{4^{k+8}}\left ( \frac{4^{k+12}.x^{k+12}}{\left ( k+12 \right )\left ( \frac{x}{4}-\frac{1}{x} \right )^{k+12}\left (4x \right )^{k+12}} +\frac{\left (2 \right )4^{k+10}.x^{k+10}}{\left (k+10 \right )\left ( \frac{x}{4}-\frac{1}{x} \right )^{k+10}\left (4x \right )^{k+10}}+\frac{4^{k+8}.x^{k+8}}{\left ( k+8 \right )\left ( \frac{x}{4}-\frac{1}{x}\right )^{k+8}\left ( 4x \right )^{k+8}}\right ){\color{Black} +c}$

$\\ =-\left ( \frac{4^{4}.x^{k+12}}{\left ( k+12 \right )\left ( x^{2}-4 \right )^{k+12}}+\frac{\left ( 2 \right )4^{2}.x^{k+10}}{\left ( k+10 \right )\left ( x^{2}-4 \right )^{k+10}}+\frac{x^{k+8}}{\left ( k+8 \right )\left ( x^{2}-4 \right )^{k+8}} \right ){\color{Black} +c} \\ \\ {\color{Blue}= \frac{-256x^{k+12}}{\left ( k+12 \right )\left ( x^{2} -4\right )^{k+12}}-\frac{32x^{k+10}}{\left ( k+10 \right )\left ( x^{2}-4 \right )^{k+10}}-\frac{x^{k+8}}{\left ( k+8 \right )\left ( x^{2} -4\right )^{k+8}}+c}$

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