Question

Evaluate:

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

Answer 1

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

$=\frac{1+\frac{1}{x^2}}{(\frac{1}{x}-x)\sqrt{x^2+\frac{1}{x^2}}}dx$

Put $x-\frac{1}{x}=u$

Now taking derivative on both sides,

$(1+\frac{1}{x^2})dx=du$

$=\int \frac{du}{-u\sqrt{u^2+2}}$

Now, put $u=\sqrt{2}\tan \left(z\right)$

Taking derivative on both sides,

$du=\sqrt{2}{\sec }^2\left(z\right)dz$

Now, our integral becomes,

$=\int \frac{\sqrt{2}{\sec }^2\left(z\right)dz}{-\sqrt{2}\tan \left(z\right)\sqrt{2}\sec \left(z\right)}$

$=\frac{-1}{\sqrt{2}}ln|cosec\left(z\right)-\cot \left(z\right)|+C$