Integration by parts is one of the two important integration methods to learn in C4 (the other is substitution ((“What about function-derivative!?” That’s a special case of substitution.)) ). In this article, I want to run through when you do it, how you do it, and why it works, just in case you’re interested.

When you use integration by parts

Integration by parts is most useful:

  • When substitution doesn’t look promising; or
  • When integrating two things multiplied together; or
  • When you have a logarithm knocking around; or
  • When you can’t think what else to do.

The last one of those doesn’t always work, but it can be worth a shot in an exam if you’re completely out of other ideas.

How you integrate by parts

To integrate by parts, you always start with a product: two things multiplied together. A typical example:

$\int xe^x\d x$

There are two things there: an $x$ and and $e^x$. You’re going to call these $u$ and $v^\prime$ (($\d v/\d x$ is more ideal, but saving space beats proper notation here)) - the question is, which way around?

The rule is, you usually want $u$ to be something that gets simpler when you differentiate it. In C4, that means:

  • If you have a $\ln$ anywhere, that will be $u$; otherwise
  • If you have an $x$ or an $x^n$ anywhere, that will be $u$.

Here, we don’t have a $\ln$, but we have an $x$, so we’ll let $u = x$ and $v^\prime = e^x$.

We need $u^\prime$, which we get by differentiating $u$, so $u^\prime = 1$; we get $v$ by integrating $v^\prime$ with respect to $x$, getting $v^\prime = e^x$. (You don’t need a constant of integration right here).

If you look in the formula book, you’ll see something like $\int u \frac{\d v}{\d x}~\d x = uv - \int v \frac{\d u}{\d x}~\d x$, which is your template for how to proceed. You started with what’s on the left hand side, so now you need to figure out the right.

$uv$ is easy enough - that’s $xe^x$. The other term is harder: $\int vu^\prime~\d x = \int (e^x)(1)~\d x$… oh, wait, it’s not actually that hard. It’s just $e^x + c$. That gives a final answer:

$\int xe^x~\d x = xe^x - (e^x + c)$

You can tidy that up a bit - there’s a factor of $e^x$ in the first two terms, and you can finagle the constant to be nicer, so you end up with $e^x(x-1) + C$. Lovely!

(If you’re not sure, you can differentiate it back using the product rule to check it gives the original integrand $xe^x$. It does.)

Another example, in less excruciating detail: let’s find $\int 4x \ln(x)~\d x$

Here, we have a $\ln$, so that’s going to be $u$; that makes $u^\prime = \frac 1x$.

$v^\prime$ is clearly $4x$, making $v = 2x^2$.

Now apply the parts formula: $\int u \frac{\d v}{\d x}~\d x = uv - \int v \frac{\d u}{\d x}~\d x$

The terms on the right hand side are $uv = 2x^2 \ln (x)$ and $\int vu^\prime~\d x = \int 2x^2 \times \frac 1x~\d x$, or - more reasonably - $\int 2x~\d x = x^2 + c$.

Putting it together, $\int x\ln(x)~\d x = 2x^2 \ln(x) - (x^2 + c)$, or (with some tidying up) $x^2(2 \ln(x) - 1) + C$. Again, you can check it using the product rule.

But where does the integration by parts formula come from?

I thought you’d never ask.

You notice how you checked both of these with the product rule? That’s not a coincidence. Let’s start there:

$\frac{\d}{\d x}(uv) = u \frac{\d v}{\d x} + v\frac{\d u}{\d x}$

If you integrate both sides with respect to $x$, you get:

$uv = \int u \frac{\d v}{\d x}~\d x + \int v \frac{\d u}{\d x}~\d x$

See where it’s going? Take the second term on the right over to the left, and you have:

$uv - \int v \frac{\d u}{\d x}~\d x = \int u \frac{\d v}{\d x}~\d x$, or

$\int uv^\prime~\d x = uv - \int vu^\prime~\d x$, as required.

See? It all fits together quite neatly.