$\int\frac{1}{x^2}dx$
For solving this we use the rule $f^m.f'$ making $f^m = x^{-2}$ thus the result is $-\frac{1}{x}+C$
My question is this:
Can I use the rule $\frac{f'}{f}$? If not, why not?
I was thinking in something like $f=x^2$ and $f'= 2x$. So it would become $\frac{1}{2x}\int\frac{2x}{x^2}dx$ and the result would be $\frac{ln(x^2)}{2x}+ C$