Students taking second semester calculus soon face a dreaded technique called “integration by parts”. The mere mention of the term might strike fear in your heart, but take a deep breath because the truth is its easier than you might think.

So what is it for? If you’re at this stage in calculus then you know that there are a set of functions that can be integrated by inspection. And you also know that there are other functions that can be easily transformed so that they can be integrated by inspection using a change of variables technique called u-substitution. Integration by parts tackles another problem: what if you have a product of two functions? Often, when you’re integrating the product of two functions this is not going to give you a situation where you can just write down the answer using the basic rules for integration or by a substitution technique. Integration by parts allows us to transform the problem into something we can solve more easily by looking at the integrand as the product of a function and the derivative of a second function. By transferring or moving the derivative from that function to the other one, we can write down an integral that can be done more easily. The formula is usually presented this way:

While this is accurate enough, once I was reading a physics book, Quantum Mechanics by a David Griffiths, and he wrote the formula for integration by parts this way:

This is a better way to write it down. Integration by parts is based on the product rule for derivatives, (fg)’=f’g+g’f, and you can see that clearly here. Also it makes it clear how to rewrite the integral on the left hand side, which is what you’re presented with in problems. This might all seem a bit mysterious at this point so lets do an example. Consider:

This integral can’t be done using any of the basic rules. In other words we can’t use the power rule. We can’t integrate the exponential because its multiplied by x, and we can’t use substitution because if we said u = 2 x, we would have du = 2dx, but that extra x in the integrand messes things up once again. Since the integrand can be viewed as the product of two functions, integration by parts is the way to tackle it.

The way to tackle it is to look and see which function to take as f, and see if its simplified by taking its derivative. Sometimes that isn’t immediately obvious, but in this case we can take:

Obviously we would get a simpler integral in this case by using this definition. So now we proceed by finding g:

This integral can be done by using substitution. We take:

And then:

Refreshing our memory, the integration by parts formula tells us that

Putting this together with our calculations we’ve done above, we get:

The second integral, on the right hand side, is easy to do:

Hence:

That’s all there is to it. In your calculus homework you’re going to start off with a few problems like this one. Do these over and over until it becomes automatic. After that, they will present you with some integrals that are a little more tricky, but the idea-applying the product rule of derivatives-is the same. In future posts we’ll show you how to do integration by parts on some more difficult integrals.

Need more help? Consider Calculus without Limits.