Why is this true? $$\int_{-\infty}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt=4\int_{0}^{1}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt$$
I know that: $$\int_{-\infty}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt=2\int_{0}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt$$ but after?
You know that $$\int_{-\infty}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt=2\int_{0}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt\tag1$$
Now we have $$2\int_{0}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt$$$$=2\int_{0}^{1}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt+2\int_{1}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt\tag2$$
Then, by setting $\frac 1t=u$, the second integral of $(2)$ will be $$2\int_{1}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt$$$$=2\int_{1}^{0}\sqrt{\frac{1}{(u^{-2}-1)^2}-\frac{(n+1)(u^{-1})^{2n}}{(u^{-1})^{2n+2}-1}}\cdot\frac{du}{-u^2}$$$$=2\int_{0}^{1}\sqrt{\frac{1}{(1-u^2)^2}-\frac{(n+1)u^{2n}}{(1-u^{2n+2})^2}}du$$ $$=2\int_{0}^{1}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt.$$