How to integrate $\int \frac{f'(x)}{f^2(x)}$?

Question

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 $-\frac{1}{f(x)}$ .

What does the numbers at the top stand for? So what's $f^1$ and $f^2$. Exponents, derivatives, … — Gono, Dec 12 '16 at 09:07
Sure that's not suppposed to be $\int \frac{f'(x) } {(f(x) )^2 }dx$? — Deepak, Dec 12 '16 at 09:07
I guess It could be a typo in the book (wouldn't be the first :/ ) — Sam, Dec 12 '16 at 09:10
But just out of interest what would $\int \frac{f^1(x)}{f^2(x)}$ have been if both are derivatives (first and second)? — Sam, Dec 12 '16 at 09:25

score 2 · Accepted Answer · answered Dec 12 '16 at 09:09

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: $$ \left(\frac1{f(x)} \right)'=f'(x) \cdot \left(-\frac{1}{f^2(x)}\right)=-\frac{f'(x)}{f^2(x)}. $$

score 1 · Answer 2 · answered Dec 12 '16 at 09:11

1

$\displaystyle \int \frac{f'(x) } {(f(x) )^2 }dx =- \frac{1}{f(x)} + c$, so is it possible your expression represents that?

answered Dec 12 '16 at 09:11

Deepak

26,801

How to integrate $\int \frac{f'(x)}{f^2(x)}$?

2 Answers2

Linked