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

I have the following integral that I must simplify by making two substitutions \begin{align} \int \frac{x^2-1}{x^2+1}\cdot \frac{1}{\sqrt{1+x^4}}dx, \end{align} one of them is \begin{align} y = \frac{x^2-1}{x^2+1}, \end{align} the other, however, I do not know which it may be and which comes first. I would appreciate your help

To do that substitution is to do$$x=\sqrt\frac{1+y}{1-y}\quad\text{and}\quad\mathrm dx=\frac1{(1-y)\sqrt{1-y^2}}\,\mathrm dy.$$Then your primitive becomes$$\int\frac y{\sqrt2\sqrt{1-y^4}}\,\mathrm dy$$and now all you have to do is $z=y^2$ and $\mathrm dz=2y\,\mathrm dy$.