# Mnemonic for Integration by Parts formula?

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.

• Product rule is what I did too, but I rarely forget the formula now that I'm teaching calculus... – user25959 Nov 11 '18 at 3:49
• $$d(uv)=udv+vdu$$ $$udv=d(uv)-vdu$$ $$\int udv=uv-\int vdu$$ How much more efficient does it need to be? – bof Nov 11 '18 at 3:54
• 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 – Isaac Caballero Nov 11 '18 at 4:11
• @IsaacCaballero: Nice! I see also that there is, "Un Día Vi Una Vaca Vestida De Uniforme" (Wikipedia). – dtcm840 Nov 11 '18 at 4:17
• I always sing in my head "udv... uv minus vdu". But really I just remember it because I know how to derive it – 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.