Question

Calculate $\int \frac{x^2-1}{x^2+1}\cdot \frac{dx}{\sqrt{1+x^4}}$.

Answer 1

Given : $\int \frac{(x^2-1)}{{(x}^2+1)}\frac{dx}{\sqrt{x^4+1}}$

Let : $x=\tan \theta$

$dx={\sec }^2\theta d\theta \frac{dx}{1+x^2}=d\theta \int \frac{(x^2-1)}{{(x}^2+1)}\frac{dx}{\sqrt{x^4+1}}=\int \frac{({\tan }^2\theta -1)d\theta }{\sqrt{{\tan }^4\theta +1}}$

$=\int \frac{({\sin }^2\theta -{\cos }^2\theta )d\theta }{\sqrt{{\sin }^4\theta +{\cos }^4\theta }}=-\int \frac{({\cos }^2\theta -{\sin }^2\theta )d\theta }{\sqrt{{({\sin }^2\theta +{\cos }^2\theta )}^2-2\sin \theta \cos \theta }}=-\int \frac{{\cos }^2\theta d\theta }{\sqrt{1-\frac{1}{2}{\sin }^{2}2\theta }}=-\int \frac{\sqrt{2}\cos 2\theta d\theta }{\sqrt{2-{\sin }^{2}2\theta }}$

Let : $z=\sin 2\theta$

$\frac{dz}{2}=\cos 2\theta d\theta -\int \frac{\sqrt{2}}{2}\frac{dz}{\sqrt{2-z^2}}=-\frac{1}{\sqrt{2}}\int \frac{dz}{\sqrt{{\left(\sqrt{2}\right)}^2-z^2}}=-\frac{1}{\sqrt{2}}{\sin }^{-1}\left(\frac{z}{\sqrt{2}}\right)=-\frac{1}{\sqrt{2}}{\sin }^{-1}\left(\frac{\sin 2\theta }{\sqrt{2}}\right)=-\frac{1}{\sqrt{2}}{\sin }^{-1}\left[\frac{2\tan \theta }{\sqrt{2}(1+{\tan }^2\theta )}\right]=-\frac{1}{\sqrt{2}}{\sin }^{-1}\left(\frac{\sqrt{2}x}{1+x^2}\right)+C$

Hence the correct answer is $-\frac{1}{\sqrt{2}}{\sin }^{-1}\left(\frac{\sqrt{2}x}{1+x^2}\right)+C$