Evaluate the integral $\int \frac{1}{x^3\sqrt{x^2+x-1}}\mathrm dx$ What substitution to use in this integral? I tried to factorize $x^2+x-1$ and use a substitution $u=\sqrt{\frac{a(x-\alpha)}{x-\beta}}$.

$$I=\int \frac{1}{x^3\sqrt{x^2+x-1}}dx$$
Euler Substitution: 
Let $x=\frac{u^2+1}{2u+1}$, $$I=2\int \frac{(2u+1)^2}{(u^2+1)^3}du$$
Partial fraction get $I=2\int \frac{4u-3}{(u^2+1)^3}du+8\int\frac{1}{(u^2+1)^2}du$
