Evaluate $\int \frac{x^2-1}{\sqrt{x^6+3x^4+x^2}}dx$

Question

Evaluate the integral $$\int \frac{x^2-1}{\sqrt{x^6+3x^4+x^2}}dx$$ Since $(x^3-x)'=x^2-1$, I tried to write $x^6+3x^4+x^2$ as a function of $(x^3-x)$.

Answer 1

Observe that $$\int\frac{x^2-1}{\sqrt{x^6+3x^4+x^2}}dx=\int\frac{1}{\sqrt{\left(x+\frac{1}{x}\right)^2 + 1}}\left(1-\frac{1}{x^2}\right)dx = \int \frac{dy}{\sqrt{1+y^2}},$$ where $y := x + x^{-1}$. I think you can take it from here.