calculus - Integrating a composition

I'm need to calculate this:

$$\int g'(x) (2g(x) - \frac{1}{g^2(x)}) dx$$

I think i have to integrate by parts, so i put:

$$dv= g'(x) dx,v=g(x)$$
$$u=2g(x) - g(x)^{-2},du=2g'(x)-(g(x)^{-2})'$$

then i apply:

$$uv-\int vdu$$
and i should integrate:
$$\int g(x)( 2g'(x)-g'(x)^{-2}$$

I'm not sure this is the right way, i think there is an easier way but how?