Okay, I thought earlier that part 3 of the above series of posts would be my last one. For some reason, this series has turned out to be a somewhat popular one when considering the fact that a big chunk of blog visitors visit it. Probably, this is due to the fact that the 2008 MIT Integration Bee is going to be held sometime soon – I have no idea exactly when – or perhaps, there are other Integration Bees that are going to be held in other colleges/universities sometime soon. If I get some more feedback/interest, then I will consider posting more results/identities/tricks on this same topic.

As you might have noticed from the title of the post, this identity isn’t related to definite integrals; it is related to indefinite integrals. Of course, since definite integrals form a “subset” of indefinite integrals, we can apply this identity to either one of them.

Let me begin by posing two problems, which I ask you to solve in your head. If you are able to do so, then you probably know the trick that is stated below, and hence you may stop reading this post; else, continue reading.

Problem 1: Evaluate \displaystyle \int e^x (\frac{x-2}{x^3}) \, dx.

Problem 2: Evaluate \displaystyle \int e^x (\ln x + \frac1{x}) \, dx.

And, here’s our identity.

(4) \displaystyle \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C,

where C is the constant of integration.

Proof: We use integration by parts. Recall, \displaystyle \int u\, dv = uv - \int v\, du, where u \equiv f(x) and v \equiv g(x). So, if we let \displaystyle v = e^x, then we have \displaystyle \int e^x f(x) \, dx = \int f(x) \, d(e^x) = f(x) e^x - \int e^x d(f'(x)), which leads us to our identity.

Now, you should be able to solve the above problems in your head in just a couple of seconds if not less.

Solution 1: \displaystyle e^x/x^2 + C.

Solution 2: e^x \ln x + C.

If you have found this particular series of posts useful, drop me a comment. Doing so will provide me the motivation to post more stuff on this topic in the near future.

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