Integration by parts is no more than a modification of the product rule for derivatives (you can see a more detailed explanation in ariels’ write-up). The final equation is ∫ u dv = u*v - ∫ v du. As you can see, after using this technique, you will still have an integral to do. The hope is that the resulting integral is simpler than the previous one. The classic example to demonstrate this is ∫x*e^{x} dx. You want to pick something for u and the rest will be dv. After choosing u and dv, v and du can be found by taking the integral of dv and the derivative of u, respectively.

∫x*e

^{x} dx u=x v=e

^{x}
du=dx dv=e

^{x} dx
Now simply

plug in for u, v, du, and dv.
x*e

^{x} - ∫e

^{x} dx
The integral of e

^{x} is not only one we are able to do, it is also very easy.
x*e

^{x} - e

^{x} + c

It is possible to choose the wrong thing for u and dv, but the idea here is to get an integral that is easier than the first, so if the second integral looks harder, it is probably a good idea to start over and choose something else. For example, what if, in the previous example, u was chosen to be e^{x} instead of x.

∫x*e

^{x} dx u=e

^{x} v=(1/2)*x

^{2}
du=e

^{x} dx dv=x dx
(1/2)x*e

^{x} - ∫(1/2)x

^{2}e

^{x} dx

This integral looks worse than the first one, and will continue to look worse if you were to keep doing integration by parts. The wrong u has clearly been picked and the problem should be started over.

The question in integration by parts is always what to pick for u. Pick the wrong thing, and you may never finish without starting over from the beginning. Choose the correct substitution and you could be done within one or two steps. How can one know what to choose? I have seen many a Calculus student (myself included) stare at an integral for hours, wondering which part to substitute for u. Fortunately, there is a lovely little mnemonic device to help the bewildered Calc student past the first, and most scary, step: L. I. A. T. E.

LOGRITHEMS
INVERSE TRIGONOMETRY FUNCTIONS
ALGEBRAIC
TRIGONOMETRY FUNCTIONS
EXPONETIALS
When looking at an integral that can be solved by integration of parts, select the logarithm for u. If there aren’t any logarithms in the problem, choose the inverse trig function; if there aren’t any inverse trig functions….etc. Simply continue down the list until you have something to substitute in for u. This mnemonic will work most of the time. How about another example to show the mnemonic really does work.

∫x*ln(x) dx u=ln(x) v=(1/2)x

^{2}
du=1/x dx dv=x dx
ln(x)*((1/2)x

^{2}) - ∫(1/2)x

^{2} * (1/x) dx
(1/2)x

^{2}*ln(x) - ∫(1/2)x dx
(1/2)x

^{2}*ln(x) – (1/4)x

^{2} + c

The first time I saw this particular problem, I thought I should set u equal to x, since it would make du=dx, thereby making the integral much easier. However, doing this would mean taking the integral of the ln(x), a much more complicated thing, which itself must be done using the integration by parts method. “How?” some may ask. “You must always have two things multiplied together to use integration by parts,” they will protest. The integral of the ln(x) does have two things multiplied together. Just watch.

∫ln(x) dx u=ln(x) v=x
du=1/x dx dv=dx
x*ln(x) - ∫x*(1/x) dx
x*ln(x) - ∫1 dx
x*ln(x) – x + c

Lots of things can be integrated using integration by parts by itself, or in combination with other techniques of symbolic integration. If you’re bored, and want to try this out yourself (or just enjoy torture), here are some more integrals that can be done just by using integration by parts.

∫x*

sin(x) dx
∫(ln(x)) / (x

^{3}) dx
∫x

^{3}*e

^{-x2} dx

Source: My college Calc II class
The last trial integral is a little tricky. You must use some creative intuitions to pick the correct values for u and dv.