Question

Integrate:
$\int \frac{\sqrt{1-x^2}+\sqrt{1+x^2}}{\sqrt{1-x^4}}dx=$

Answer 1

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

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

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

$I=\log \left|x+\sqrt{x^2+1}\right|+{\sin }^{-1}+C$

Hence, this is the answer.