# Equivalent of product rule for integrals: $\int f'(x)g'(x)dx$

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 :)

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

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