$\int \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x$ is equal to

$\frac{1}{\sqrt{2}} \sin ^{-1}\left\{\frac{\sqrt{2} x}{x^2+1}\right\}$

Let

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

$\Rightarrow I =-\int \frac{1}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-(\sqrt{2})^2}} d\left(x+\frac{1}{x}\right)$

$\Rightarrow I=\frac{1}{\sqrt{2}} cosec^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)+C=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{\sqrt{2} x}{x^2+1}\right)+C$