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Question Number 108605 by bemath last updated on 18/Aug/20

$$\frac{\mathit{B}e\mathit{M}ath}{\approxeq }$$$$\int \frac{x^{11}}{{(x^8+1)}^2}dx$$

Answered by bemath last updated on 18/Aug/20

$$let{x}^4=\tan v\Rightarrow 4x^{3}dx=\sec {}^{2}vdv$$$$I=\frac{1}{4}\int \frac{x^8\left(4x^{3}dx\right)}{{(x^8+1)}^2}=\frac{1}{4}\int \frac{\tan {}^{2}v\sec {}^{2}vdv}{\sec {}^{4}v}$$$$I=\frac{1}{4}\int \frac{\tan {}^{2}v}{\sec {}^{2}v}dv=\frac{1}{4}\int \sin {}^{2}vdv$$$$I=\frac{1}{8}\int (1-\cos 2v)dv$$$$I=\frac{1}{8}v-\frac{1}{8}\sin v\cos v+c$$$$I=\frac{1}{8}[{\tan }^{-1}\left(x^4\right)-\frac{x^4}{x^8+1}]+c$$

Answered by malwan last updated on 18/Aug/20

$$y=x^4\Rightarrow dy=4x^{3}dx$$$$\Rightarrow \int \frac{x^8\times x^3}{x^8+1}dx=\frac{1}{4}\int \frac{y^2}{y^2+1}dy$$$$=\frac{1}{4}\int \frac{y^2+1-1}{y^2+1}dy$$$$=\frac{1}{4}\int [1-\frac{1}{y^2+1}]dy$$$$=\frac{1}{4}[y-{tan}^{-1}y]+c$$$$=\frac{1}{4}[x^4-{tan}^{-1}{x}^4]+c$$

Commented by bemath last updated on 18/Aug/20

$$cooll$$

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