EJK's Random Calculus Tutorial
This is a page I decided to write about tabular integration and how to possibly make tabular integration work if you have a yucky product with nothing that can be derived to zero. I'm assuming that you already know integration by parts (u dv = u * v - v du). What I'm describing here is a way to sidestep the tedium of reapplying parts to v du every time.

First, how tabular integration ideally works: You have a yucky product of functions, and can find one thing that can eventually be derived to zero. Example:

x^2 * sin(x) dx
u = x^2dv = sin(x)
du = 2xv = -cos(x)
d^2u = 2v = -sin(x)
d^3u = 0v = cos(x)
The solution is u*v - du*v + d^2u * v ... [+-?] d^Nu * [N]v, where the Nth derivative of u is a constant times N integrals of v.

Look and see how this is an automation of integration by parts. The solution is u * v - v du. That integral is, in turn, the derivative of u times the integral of v minus the integral of v times the second derivative of u. Rinse, lather, repeat - eventually, u goes to zero and you're left with some nth integral of v times nothing. You now have a very quick way of integrating some mess if you can find a u that will derive to zero (provided that v isn't too evil to integrate, that is).

Now suppose you have some monstrosity where you have TWO parts and neither of them can be derived to zero (which is the point of this page). It might be hopeless, or it might not be. If both parts are functions whose derivatives and/or integrals make repeating sequences (for example, sin -> cos -> -sin -> -cos -> sin ...; e^x -> e^x forever; integrals of ln(x) -> 1/x -> ln(x)...) then it may be possible to use tabular integration. The key is that the nth integral of dv times the nth derivative of u times a constant equal the original u and dv. The exception is that the constant must not be -1, because then adding the integral to both sides would eliminate it. Example:

e^x * sin(x) dx
u = e^xdv = sin(x)
du = e^x>v = -cos(x)
d^2u = e^xv = -sin(x)
d^3u = e^xv = cos(x)
d^4u = e^xv = sin(x)
Obviously, this table never ends - the derivative of e^x is e^x no matter what. But there IS hope! On the 3rd row, we have a constant times the original integral. Stop at the row above the matching row. Since we're going to subtract du*v, we have - [-e^x*sin(x) - -e^x*sin(x)]. This reduces to e^x*sin(x) - e^x * sin(x). So we now have:

e^x * sin(x) dx = -e^x * cos(x) + e^x * sin(x) - e^x * sin(x) dx.

2e^x * sin(x) dx = -e^x * cos(x) + e^x * sin(x)

e^x *sin(x) dx = ((-e^x * cos(x) + e^x * sin(x)) / 2) + C

I confirmed that that is the correct integral by using my TI-83's integrate function over 5 to 8 and -3 to 1 and plugging those numbers into my own function. It is indeed correct. If you want to derive it to get the original function... I honestly don't have a clue how to make e^x * sin(x) stay e^x * sin(x).

Now I'm gonna try something ugly. I may have a mistake in here - correct me if I do! ln(2x) * cos(x/3) dx
u = cos(x/3)dv = ln(2x)
du = -sin(x/3)/3v = 1/2 * 1/2x
d^2u = -cos(x/3)/9v = ln(2x)/4
d^3u = sin(x/3)/27v = 1/8 * 1/2x
Not as pretty, but manageable. On the 3rd line again, we have a constant (-1/36) times the integral of what we were originally trying to integrate. Now it's just simplifying:

ln(2x) * cos(x/3) dx = cos(x/3)/4x - [ -sin(x/3)*ln(2x)/12 - -cos(x/3)*ln(2x)/36 ]

ln(2x) * cox(x/3) dx = cos(x/3)/4x + sin(x/3)*ln(2x)/12 + 1/36cos(x/3)*ln(2x)

35/36ln(2x) * cos(x/3) dx = cos(x/3)/4x + sin(x/3)*ln(2x)/12

ln(2x) * cos(x/3) dx = (36/35) * (cos(x/3)/4x + sin(x/3)*ln(2x)/12) + C

Ugly, but none the less manageable.