Question

Solve:$\int \frac{dx}{(x^2+1{)}^3}$

Answer 1

$\int \frac{dx}{{(x^2+1)}^3}Letx=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta \therefore \int \frac{dx}{{(x^2+1)}^3}=\int \frac{{\sec }^2\theta d\theta }{{({\tan }^2\theta +1)}^3}=\int \frac{1}{{\sec }^4\theta }d\theta (\because 1+{\tan }^{2}x={\sec }^{2}x)=\int {\cos }^4\theta d\theta =\int {\left({\cos }^2\theta \right)}^{2}d\theta =\int {\left(\frac{\cos 2\theta +1}{2}\right)}^{2}d\theta (\because \cos 2x=2{\cos }^{2}x-1)=\frac{1}{4}\int {(\cos }^{2}2\theta +2\cos 2\theta +1)d\theta$

$=\frac{1}{4}\int [\frac{\cos 4\theta +1}{2}+2\cos 2\theta +1]d\theta =\frac{1}{8}\int [\cos 4\theta +4\cos 2\theta +3]d\theta =\frac{1}{32}[\sin 4\theta +8\sin 2\theta +12\theta ]+C=\frac{1}{32}\left[\sin \right(4{\tan }^{-1}x)+8\sin (2{\tan }^{-1}x)+12({\tan }^{-1}x\left)\right]+C$