The Integration by Parts formula may be stated as: $$\int uv' = uv - \int u'v.$$

I wonder if anyone has a clever mnemonic for the above formula.

What I often do is to derive it from the Product Rule (for differentiation), but this isn't very efficient.

One mnemonic I have come across is "ultraviolet voodoo", which works well if we instead write the formula as: $$\int u \ \textrm{d}v = uv - \int v \ \textrm{d}u.$$

I am however looking for a mnemonic for the first formula.

  • $\begingroup$ Product rule is what I did too, but I rarely forget the formula now that I'm teaching calculus... $\endgroup$ – user25959 Nov 11 '18 at 3:49
  • 4
    $\begingroup$ $$d(uv)=udv+vdu$$ $$udv=d(uv)-vdu$$ $$\int udv=uv-\int vdu$$ How much more efficient does it need to be? $\endgroup$ – bof Nov 11 '18 at 3:54
  • 2
    $\begingroup$ In spanish we have "una vaca menos una vaca vestida de uniforme", but I completely agree with using the product rule is one of the best suggestions since you can truly understand IBP $\endgroup$ – Isaac Caballero Nov 11 '18 at 4:11
  • $\begingroup$ @IsaacCaballero: Nice! I see also that there is, "Un Día Vi Una Vaca Vestida De Uniforme" (Wikipedia). $\endgroup$ – dtcm840 Nov 11 '18 at 4:17
  • $\begingroup$ I always sing in my head "udv... uv minus vdu". But really I just remember it because I know how to derive it $\endgroup$ – clathratus Nov 12 '18 at 0:15

There are a couple of ways that you can remember things like this, I always did it by writing it out in different ways and seeing which one worked for me: $$\int uv'=\int u\frac{dv}{dx}dx=\int udv$$ I know some people preferred different forms of the notation above (whether strictly correct or not) so I have included them all.

By writing out your know 'values' in matrix form you can work out what is left by seeing which diagonal has not been done, as I show below: $$\begin{matrix} u&v'\\ u'&v \end{matrix}$$ The first integral (that we are trying to solve) is the horizontal at the top $uv'$. we now want to deal with the diagonal and the bottom horizontal, $uv$ and $u'v$. I'd remember that since $uv$ has no prime, we do not integrate, and since $u'v$ has a prime we do integrate. Now we minus the integral from the part that is not integrated.

My wording is not the best but if you write out the notation hopefully you will understand what I mean


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.