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Recently I've been working on some integration problems and I have come across some integrands of the form $f'(x)g'(x)$. I've found myself wondering; I know the product rule for differentiation, written below: $$\frac{d}{dx}f(x)g(x)=f'(x)g(x)+g'(x)f(x)$$ but is there a similar formula for the integral below? (I have written $f'(x)g'(x)$ as opposed to $f(x)g(x)$ as I'm talking about a case when I know the integral of each individual function in the integrand.) $$\int f'(x)g'(x)dx$$ In order that you don't give me answers that are too advanced for me, I'll quickly list what I have covered in calculus so far: differentiation (chain,product, basic,quotient,$\ln$, exponential,trig etc); integration using reverse chain rule, substitution,separation of variables, 1st and 2nd order differential equations in both $x$ and $y$, use of partial fractions,integration by parts,trig integration, some basic hyperbolic functions,exponentials and logarithms. Please comment if you would like more information.

Thnak you for your help, it is much appreciated :)

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  • $\begingroup$ no i dont think so one can use integration by parts to get the best $\endgroup$ Sep 2 '20 at 12:51
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    $\begingroup$ That would be every A level student's dream....but in the real world we just have integration by parts. $\endgroup$ Sep 2 '20 at 12:53
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How about this

$$(fg)'' = f'' g + 2f' g' + g''$$

Thus

$$\int f'g' = \int \underbrace{\frac{f''g}{2} + f'g' + \frac{g''f}{2}}_{\frac{(fg)''}{2}} - \frac{f''g}{2} - \frac{g''f}{2} = \frac{1}{2} (fg)' - \frac{1}{2}\int f''g-\frac{1}{2}\int g''f$$

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    $\begingroup$ Very nice the answer. $\endgroup$
    – Sebastiano
    Sep 2 '20 at 14:29

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