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I have a curiosity. If

$\int f(x) f'(x)dx=\int f(x) df(x)=\frac{\left(f(x)\right)^{2}}{2}+C$

what is the result of:

$\int f(x) f''(x)dx$

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Yes. And $\int (f'(x))^2 dx$? – Mark Jul 4 '12 at 10:59
To write the solution without an integral is big and deep problem for mathematics. – Mathlover Jul 4 '12 at 11:00
@Mathlover, reference please? – lhf Jul 4 '12 at 11:05
@lhf:I do not know any technics to solve $\int f'(x)^2 dx $ without endless series. if you have a closed form solution of $\int f'(x)^2 dx$ and also you would have a closed form of $\int 4x^2e^{2x^2} dx$ when $f(x)=e^{x^2}$ with elemantary functions. They are related to each other. Please see – Mathlover Jul 4 '12 at 11:40
@Mathlover, that's a nice observation, thanks. – lhf Jul 4 '12 at 11:57
up vote 2 down vote accepted

$$ f(x)f'(x)-\int f'(x)^2 \, dx \ ? $$

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Yes. And $\int (f'(x))^2 dx$? – Mark Jul 4 '12 at 11:01
In general, nothing: it is what it is. – Siminore Jul 4 '12 at 11:02
This isn't Jeopardy! – draks ... Jul 4 '12 at 11:03
@draks What is math.stackexchange? – anon Jul 4 '12 at 11:07

Using by parts $\int f(x)f''(x)dx= f(x)f'(x)-\int f'^2(x)dx$.

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