How do you do integration by parts...is there a good example problem you can do to prepare for a test?

Answer 1

Integration by parts is a technique used when you want to take the intregral of a function that is actually two smaller functions (more easily integratable/differentiable) multiplied together. For example, the integral of x*sin(x).

Here's the formula for integration by parts (just pretend my integral sign | is more squiggly): |u*dv = uv - |v*du .

The left side of the equation should be matched up to the problem you are given. Using the example above, x will be playing the role of u, and sin(x) will be playing the role of dv (you could interchange them, but I think this way's easier).

So you've matched up your problem with the left side of the equation. You can use the rest of the equation to rewrite this same integral as "something minus an integral that's easier to solve." Your ingrediants: the u you've picked, its derivative (du), the dv you've picked, and its antiderivative (v).

It helps to write it out:

u = x

du = 1

v = -cos(x)

dv = sin(x)

Using the equation, then, we can rewrite the complicated integral that we split apart:

|u*dv = uv - |v*du

|x * sin(x) = -x*cos(x) - |-cos(x)*1

and allow me to simplify the right side of the equation:

= -x*cos(x) - |-cos(x)

= -x*cos(x) + |cos(x)

= -x*cos(x) + sin(x)

and always, always, remember to add a possible constant once you have taken an indefinite integral; so, drum roll, please:

|x * sin(x) = -x*cos(x) + sin(x) + C. This looks complicated, but is "solved" because it includes no integral signs | and you cannot combine any like terms.

*Here's a practice problem for you: find | ln(x) dx. Recall that the ln means "the natural log of."

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Hint #1: Rewrite it as |ln(x) * 1 dx.

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Hint#2: Choose u = ln(x) and dv = 1, because you'll have to take the antiderivative of dv.

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Hint #3: Don't forget to add +C!!!

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Answer = x*ln(x) - x + C.