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I am trying to seek a close form of \begin{eqnarray*} \int \frac{f''(x)f'(x)}{f(x)}\mathrm{d}x \end{eqnarray*} can you help me guys?

This is my first attemp: \begin{eqnarray*} \int \frac{f''(x)f'(x)}{f(x)}\mathrm{d}x&=&\int\frac{y}{Y}\mathrm{d}y=\ln Y+c=\ln f(x)+c \end{eqnarray*} which is obviously wrong, but I don't know where and how to fix it. ($y=f'(x)$ and $Y=\int y\mathrm{d}y$)

My second attemp: \begin{eqnarray*} \int \frac{f''(x)f'(x)}{f(x)}\mathrm{d}x=\int\frac{(\mathrm{d}y)^2}{y} \end{eqnarray*} which is impossible(?) ($y=f(x)$).

Thank you so much in advance, I appreciate your help.

Rizky Reza Fujisaki
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2 Answers2

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By parts,

$$\int\frac{f''(x)f'(x)}{f(x)}dx=\frac{f'^2(x)}{2f(x)}+\frac12\int f'^2(x)\frac{f'(x)}{f^2(x)}dx.$$

Then there is nothing you can simplify in

$$\int\frac{f'^3(x)}{f^2(x)}dx.$$

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Note that another simplification (seriously?), in addition to Yves's one can be to identify that: the derivative of $\ln f (x) $ is $$\frac {f'(x)}{f (x)} $$ and thus integrating by parts will give us: $$\int f''(x) \mathrm d (\ln f (x)) = f''(x)\ln f (x) - \int \ln f (x) \times f'''(x)\, dx $$

Now, there us nothing we can do to simplify the last integral.