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 :)

  • $\begingroup$ no i dont think so one can use integration by parts to get the best $\endgroup$ Sep 2 '20 at 12:51
  • 3
    $\begingroup$ 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

How about this

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


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

  • 1
    $\begingroup$ Very nice the answer. $\endgroup$
    – Sebastiano
    Sep 2 '20 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.