$\int \frac{1+x^4}{\left(1-x^4\right)^{3 / 2}} d x$ is equal to

$\frac{1}{\sqrt{\frac{1}{x^2}-x^2}}+C$

Let

$I=\int \frac{1+x^4}{\left(1-x^4\right)^{3 / 2}} d x=\int \frac{\frac{1}{x^3}+x}{\left(\frac{1}{x^2}-x^2\right)^{3 / 2}} d x$

$\Rightarrow I=-\frac{1}{2} \int \frac{-2 x-\frac{2}{x^3}}{\left(\frac{1}{x^2}-x^2\right)^{3 / 2}} d x=-\frac{1}{2} \int \frac{1}{\left(\frac{1}{x^2}-x^2\right)^{3 / 2}} d\left(\frac{1}{\left.x^2-x^2\right)}\right.$

$\Rightarrow I=-\frac{1}{2} \times \frac{\left(\frac{1}{x^2}-x^2\right)^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+C=\left(\frac{1}{x^2}-x^2\right)^{-1 / 2}+C$

$\Rightarrow I=\frac{1}{\sqrt{\frac{1}{x^2}-x^2}}+C$