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Even though I've been through countless instances where I needed to use integration by parts, to this day, I still derive it from the chain rule, identify my 'parts' and apply the formula.

At some lecture a while back, one of my profs did a two step integration by parts in like 30 seconds. It went so fast, looked so easy, and I never caught on to how he did it. The answer just fell out.

What are you favorite tricks for quick integration by parts? Everything goes, mnemonics, special notation, or maybe some other trickery?

Speed is the number one priority here. I'd rather have a fast method that requires some practice to master, than have a slow method that can be learnt quickly!

Thanks in advance!

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  • $\begingroup$ You mean if you saw $\int x \sin x \ dx$ you would first rederive integration by parts for this case via the product rule of differentiation? I suggest practicing a lot more to do it without reverting back to 'first principles'. $\endgroup$ Commented Jun 18, 2015 at 19:39
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    $\begingroup$ I think you're looking for this: tabular integration by parts. $\endgroup$ Commented Jun 18, 2015 at 19:45
  • $\begingroup$ No, i mean if I see that integral, I would write $ d(ux) = udx + xdu$, integrate, rewrite, write something like $u = ...$ and $v = ...$, plug in and go. Isnt this a bit slow? $\endgroup$ Commented Jun 18, 2015 at 19:45

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The main result is $$\int udv = uv - \int vdu$$ which means you pick out 2 parts from your integral -- you will differentiate one, and integrate the other one, which often will give you some idea of what the next integral will look like before any arithmetic was carried out.

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  • $\begingroup$ I think I was expecting something more miraculous. I don't like the tabular method that John describes, since it seems to hide the mathematical machinery, and most people seem to have the formula memorized... Maybe I just have a weak memory, who knows.... I think I'll do one integration by parts problem every morning for a month or so, it'll stick eventually.... $\endgroup$ Commented Jun 18, 2015 at 21:46
  • $\begingroup$ I'll leave this answer unaccepted, just in case someone really is holding onto some secret way that I've never seen before $\endgroup$ Commented Jun 18, 2015 at 21:48
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It might have been the Stand and Deliver "tic-tac-toe" method.

It involves a table: derivatives of one function, anti-derivatives of the other function, and a column with an alternating sign.

So things like $\int x^5 \cos(x) dx$ are really fast. Take derivatives of $x^5 (5x^4, 20x^3, 60x^2, 120x, 120)$, anti-derivatives of $\cos(x) [\sin(x), -\cos(x), -\sin(x), \cos(x), \sin(x), -\cos(x)]$. Multiply the $x^5$ by the first anti-derivative in the list, and keep the sign ($x^5 \sin(x)$). Multiply the $5x^4$ by the second anti-derivative, and switch the sign ($5x^4 \cos(x)$). Continue until the term is zero:

$$\int x^5 \cos(x) dx \\ = x^5 \sin(x)+5 x^4 \cos(x)-20 x^3 \sin(x)-60 x^2 \cos(x)+120 x \sin(x)+120 \cos(x)+C.$$

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  • $\begingroup$ The tabular (tableau) method $\endgroup$ Commented Jun 18, 2015 at 20:42

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