Its always a good idea to start out writing the integration by parts formula:

For now, we can simply ignore the limits of integration. First we take

Then we have

And so

Now we’ve got to apply integration by parts on the integral we have at the end. Let’s go through it in case you aren’t familiar with it. Take

Hence

Now we can put this together with our previous result, and we’ve found that

If we were doing an indefinite integral, we would add a constant of integration to the end and we’d be done. For integration by parts with limits, all we do now is evaluate this expression at the limits of integration. The upper limit was x = 3, which gives:

The lower limit was x =2, which gives

So the answer is found by subtracting the lower limit from the upper limit, and you can verify that

So, to do integration by parts with limits, simply do the integral the way you would in the indefinite case, ignoring the limits of integration. Then when you have your answer at the end, evaluate at each limit and subtract.

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Its always a good idea to start out writing the integration by parts formula:

Since taking the derivative of x would just give 1, we’ll take that to be u. So

This integral must be done using u-substitution. Let’s go ahead and show the details for readers who aren’t sure how to proceed. Take

Then we have, ignoring the constant of integration that we’ll tack on at the end of the problem

Putting back the original variable

Now let’s put this result together with the integration by parts formula to solve the original problem. Now we have

The last integral can be done using u-substitution again. We won’t go through the details this time. Adding the constant of integration it is

So the answer is

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First let’s recall the integration by parts formula:

The trick is to pick u and v in such a way that

is a simpler expression. In this case, try

Then

(ignore constants of integration for now, we’ll add it at the end). And so

It seems that we haven’t gotten anywhere. But we can apply integration by parts again on the second integral. We take

Hence

Putting everything together

Now add

To both sides. Then we find (making sure to add in the constant of integration)

Let’s remind ourselves how to do integration by parts:

At first glance, we might be tempted to take:

But if we did that, we would be faced with the problem of finding:

So we’ve hit a wall. Instead, we will approach the problem by taking the natural logarithm to be u, and then we’ve got:

So the integral becomes:

Evaluating the integral we get the final result:

This example illustrates that the right choice of u and dv can save a bit of time and effort when solving integration by parts problems in calculus.

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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.

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