How to integrate $\int \frac{f'(x)}{f^2(x)}$? In my book it states $\int \frac{f^1(x)}{f^2(x)}=-\frac{1}{f(x)}$.
I don't really understand why and it doesn't say in my book.
My first idea was maybe we can cancle out $f^1(x)$. So  $\int \frac{f^1(x)}{f^2(x)} = \int \frac{1}{f^1(x)}$.
But I'm not sure if you can do that and it also doesn't explain the minus from $-\frac{1}{f(x)}$ .

Hint. Let's assume it's a derivative on the top. This comes from the fact that
$$
\left(\frac1{x} \right)'=-\frac1{x^2}
$$ and from using the chain rule:
$$
\left(\frac1{f(x)} \right)'=f'(x) \cdot \left(-\frac{1}{f^2(x)}\right)=-\frac{f'(x)}{f^2(x)}.
$$
