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?
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