We learn a new technique, called integration by parts, to help find antiderivatives of certain types of products by reexamining the product rule for differentiation.

Consider the following example.

If we use the shell method to set up an integral that gives the volume of , we obtain since and .

To determine the volume of we need to find an antiderivative of . However, none of our previous methods allow us to do this.

In order to develop a technique that allows us to find the desired antiderivative, remember that whenever we have a formula that allows us to differentiate a function , there is a corresponding integration result. We notice that the above integrand involves a product, so let’s try starting with the product rule for derivatives and try to reverse it.

First note that by the product rule we have for differentiable functions and Now integrate both sides of the equation above By the Fundamental Theorem of Calculus, we can handle the lefthand side of the equation. Also, by the linearity of integrals we can rewrite the righthand side as Hence, we find and so by a little rearranging,

Now, let’s return to the volume example, where we found

We first find the antiderivatives of and make the seemingly bold claim that the previously developed formula will help us find them. Here’s how it works.

Since is a product and none of our previous techniques work, we can try integration by parts. Note that the formula requires that we pick part of the product to differentiate and part of it to integrate. Since becomes constant when we differentiate it, and the antiderivatives of are easy to find, let’s try and .

The integration by parts result gives us the following.

To ensure that this result is correct, we can check by differentiating the righthand side and comparing it to the original integral.

Thus, we have found the antiderivatives for , and we can use them along with the Fundamental Theorem of Calculus to find the volume of .

Now, the integration by parts formula for the indefinite integral would yield the following.

Note that the term is the constant associated with the last integral on the righthand side. The constant that we included when integrating the term cancels! This is not an accident, and it is not hard to generalize this argument to conclude that

we do not need to include a constant of integration when we antidifferentiate .

### The integration by parts formula

The technique we applied in the above example extends to many other *products*
of functions. We summarize the result of the argument in the preceding
example first, then study other types of integrals for which this is a useful
technique:

In general, we want to choose so the resulting integral on the righthand side is “easier” to compute, but in doing so, we must be able to integrate our choice for .

We’ll now work some standard examples to develop some intuition for the technique.

Note that if we choose , and the righthand side of the integration by parts formula will no longer include a polynomial. This choice will work if we can integrate the other part of the product in the original integral. Indeed, if we set , we must choose , and since this is not difficult to antidifferentiate, let’s give it a try.

using the integration by parts result gives the following.

We once again can check the validity of the proposed antiderivative above by computing the derivative of our proposed answer.

*differentials*and since and are functions of , the differentials and must include a term.

Sometimes, we have “disguised products” in the integrand, as in the following example.

and so by using the integration by parts formula, we obtain the following.

Sometimes, several techniques arise in the computation of antiderivatives.

We still need to determine , but note that a substitution is helpful. In fact, let . Then and we find the following.

Putting it all together gives You should check for yourself that if we differentiate the answer above, we obtain the original integrand.

### Repeated integration by parts

The integration by parts formula is intended to replace the original integral with one that is easier to determine. However the integral that results may also require integration by parts. This can lead to situations where we may need to apply integration by parts repeatedly until we obtain an integral which we know how to compute.

**tabular method**that allows for more efficient computation. We illustrate with the previous example. We form a table by successively differentiating the entries in the column for and antidifferentiating the entries in the column for . In the “sign” column, we start with a and alternate. Here is the table.

To write down the antiderivative, compute with this second table we begin at the top.

- Step 1: Include the sign with the first entry for , and multiply the result by the second entry in column .
- Step 2: Include the corresponding sign with the entry , and multiply the result by the entry in column one row below the current one for .
- Step 3: If you have obtained 0 in the previous calculation, stop this procedure. Otherwise, turn your attention to the next lowest row and continue performing the process from Step 2 until you arrive at 0. Don’t forget the constant of integration at the end.

Here’s how it works in the chosen example. Note that the explicit results are given inside of the square brackets.

Sometimes, we can perform integration by parts a few times and recover our original integral.

Choose and . Then we have Using the integration by parts result, we find

The integral on the right looks very similar to our original integral. But we persist and try integration by parts on the rightmost integral .

Choose and . This gives us

and so

Plugging this into the formula we derived previously, we have

It may appear that we have not made any progress. We have applied integration by parts twice and all we have managed to do is obtain our original integral again.

To see how this is helpful, denote our original integral by , so . The above formula we derived can be written as

which is an algebraic equation in that can be solved. Indeed, we find

and use this to conclude that

Note we need to include an arbitrary constant of integration since we are finding an antiderivative. The above technique is a useful one to keep in mind. If you do integration by parts multiple times and seem to be getting nowhere, check to see if the original integral reappears. If it does, then it’s possible solve for the integral in the same manner as we did in this example.

Sometimes we can use integration by parts to give a **reduction formula**.

This reduction formula shows how to ”reduce” the power of so the resulting integral is slightly easier. We may use repeated applications of this rule to keep reducing the power of until we arrive at an integral we know how to compute directly.

### Final thoughts

When we covered the substitution method for antiderivatives, we saw that there was no fixed procedure for choosing . There were only certain rules of thumb that might guide you to better or worse choices of which part of the integrand to substitute. The same idea applies in integration by parts. There is no procedure that tells you the best choice for and . However here is a useful heuristic (“rule of thumb”) that can guide your choice.

The heuristic is referred by the mnemonic *ILATE*. The individual letters stand for
different types of functions:

The idea is that when choosing and , one looks at the types of functions that show
up in the integrand. Function types that occur earlier in *ILATE* are better choices
for while those that appear late are better choices for . This is because
functions near the top of *ILATE* generally become simpler when they are
differentiated.

For example, in , the integrand is made up of a product of a algebraic function and a
logarithmic function . ILATE would suggest choosing since logarithmic comes above
algebraic in *ILATE*. Thus since that is the remaining portion. However one should
keep in mind that *ILATE* is simply a rule of thumb that does not always
apply and can actually make the problem more difficult to solve in some
instances.

“Mathematics is the science of skillful operations with concepts and rules invented for this purpose” - Eugene Wigner