When it is not possible to transform the integral to be able to use the immediate integral, we use the method of integration in parts, which is what we will see in this lesson.

Let’s go!

Índice de Contenidos

## Integration by parts

The piecemeal method is based on the following formula:

It is used when it is not possible to integrate by means of the immediate integrals, since it is not possible to transform the integral so that it resembles some of its formulas.

Note, that in the solution of the formula of integration by parts there is another integral one. That integral can be solved again in parts or by means of immediate integral. The process is iterative, i.e. it has to be repeated over and over again.

If it is resolved again in parts, we must see if the integral is simpler than the original and therefore we are on the right track and we must continue iterating the method until we reach an integral that is resolved by immediate integral and therefore we arrive at a solution without integral.

To integrate by parts we must identify within the integral a function f’, that will have to integrate and a function g that will have to derive:

Once we have integrated f and derivative g, we have the 4 functions we need to apply the integration formula in parts and we only have to replace them.

All this will be easier when we begin to solve examples of integration in parts.

## How to choose the function to be integrated and the function to be derived in the integration method in parts

Before starting to solve integral parts, I’m going to leave you some indications so that you have some criteria when choosing which function to integrate and which to derive.

How to choose f’ and g?

Normally, you choose f’ the function that is easiest to integrate and leave the g to derive, since deriving a function is easier than integrating it.

They are good candidates to be the function to integrate f’:

- Function of the number e elevated to x
- Trigonometric functions: cos x, sen x…
- Potential functions: x², x³, etc…
- El propio dx

When several of these functions coincide in the integral, it always chooses the function to integrate in the order of priority in which they are in the previous list, that is, if we have this integral:

The function to integrate will be that of e elevated to x and the function cos x will be the function to derive:

The differential of x dx, must always be in the derived function.

On the other hand, the function ln x can never be f’, because to integrate this function, the method of integration in parts has to be used.

The inverse trigonometric functions (arc sen, arc cos, arc tag) cannot be the function to integrate f’ either.

What if I don’t choose right f’ and g?

If you do not choose well the function to integrate and to derive, each time the integral will be more complicated, therefore, you must stop and change your choice.

Remember that your objective is to reach an integration that can be solved by immediate integrals and that therefore allows you to reach the integral solution.

## Resolved integration exercises by parts

So far, everything I’ve told you may be difficult for you to assimilate, but don’t worry. We are going to settle concepts solving a few integrals by the method of integration in parts, in which I will explain each of the steps and you will understand better the operation of this method.

I recommend that once you finish studying the exercises, you read the previous theory again so that you understand everything much better.

All these integrals have one thing in common and that is that it is not possible to solve them with the immediate integrals method, since they do not have the same form as any of them nor can they be retouched to have it.

Therefore, we will start from that base. Keep that in mind.

### Exercise resolved 1

First, we pull out the constant:

We have to choose one function to integrate and another to derive. As a function to integrate, we choose cos x, since it is above in the priority list. We leave as a function to derive the x:

We integrate f’ and derive g:

Note that dx always goes with derivatives.

Now we have everything we need (f, g, f’ and g’) to apply the formula of integration in parts:

So we replace them with their corresponding functions:

We have yet to resolve an integral, which is resolved by immediate integral:

Finally we operate and add the constant:

### Exercise resolved 2

We choose the function to be derived and the function to be integrated:

In this case, it is simple, because ln x can never be the function to be integrated, so there is no choice but to derive the function.

We integrate f’ and derive g:

We apply the formula of integration in parts:

In the integral that we have left, we take out the constant and we operate:

We are left with an immediate integral that we will go on to resolve:

Finally, we operate on the second term and add the constant:

### Exercise solved 3

In this case, we have two functions candidate to be the function to integrate. We choose cos x to be higher in the order of priority and leave the other function to derive:

We integrate f’ and derive g:

We apply the formula of integration in parts:

In the integral that we have left, we take out the constant:

We have been left with an integral that cannot be solved through immediate integral, so we have to solve it again in parts. We follow the same criteria as before to choose the function to be integrated and the function to be derived:

We integrate f’ and derive g:

We again applied the formula of integration in parts:

We take out of the integral the minus sign, which is equivalent to having a -1 and therefore is a constant:

This time the integral that we have left can be solved by immediate integral, so we go on to solve it:

We operate to remove the brackets and add the constant C:

### Exercise solved 4

At first glance, this integral may seem like it could be solved with an immediate integral, but it is not.

It has to be solved in parts, but it seems to have only one function and that can cause confusion. However, dx can also be used as a function to be integrated or derived since it corresponds to the function derived from x.

Therefore, we chose dx as a function to be integrated and ln x as a function to be derived. In addition, everything fits together because ln x cannot be selected for the function to be integrated:

We integrate f’ and derive g:

We apply the formula of integration in parts:

We operate within the integral that we have and we have left:

We solve the integral and add the constant:

### Exercise completed 5

This integral is a similar case to the previous one. We choose as function to integrate the differential of x and the arc sen x as function to derive:

We integrate f’ and derive g:

We apply the formula of integration in parts:

The integral that remains is resolved by means of the immediate integral of the compound potential function, since we have a function elevated to an exponent (the root) and we could have the derivative of that function, where the function is:

And its derivative:

We add the missing 2 so that the derivative appears and divide the integral by 2:

We now apply the formula of the immediate integral of the compound potential function:

We operate on the second term and add the constant:

## Cyclic integrals in the integration by parts

As you have seen in the previous exercises, as the method of partial integration is an iterative process, it is sometimes necessary to reapply the method in the new integrals that arise.

It may be the case that the new integral that emerges is the same as the original integral, so if we continue applying the same procedure, we would enter into an infinite cyclic process and we would never finish solving it.

Let’s see an example of these integrals at the same time that I explain the procedure to solve them:

We choose the function to integrate the trigonometric function and the exponential function as the function to derive:

We integrate f’ and derive g:

We apply the formula of integration in parts:

We take out of the new integral the ln 2 for being a constant:

The remaining integral is solved again in parts. We follow the same criteria for choosing f’ and g:

We integrate f’ and derive g:

We apply again the formula of integration by parts in the new integral:

We take out the integral that’s left of ln 2:

And if you realize at this point of the exercise, I repeat the previous step but highlighting the integral that we have left:

Ring a bell?

Indeed, it’s the original integral we’re solving.

What do we do now?

If we stay the same, we’ll never finish, but there is a solution.

We are going to call the original integral I:

Therefore, all the development we have so far is equal to I and in addition, the integral that appears within the development is also equal to I, so we replace it with this variable:

Now what we have is an equation where the unknown is I and we have to clear it.

First, we multiply ln 2 by the terms inside the bracket to remove it:

Now we pass the term with I adding the first member:

In the first member, we take common factor to the I:

And finally, we pass dividing the content of the bracket that mutiplica to the I and add the constant:

As you can see, in order to know when the split integration method is applied, you must identify one possible function to integrate and another to derive, as long as it is not possible to solve it with immediate integrals.