Antiderivatives can be difficult enough to solve on their own, but when you’ve got two functions multiplied together that you need to take the antiderivative of, it can be difficult to know where to start.

That’s where the integration by parts formula comes in!

This handy formula can make your calculus homework much easier by helping you find antiderivatives that otherwise would be difficult and time consuming to work out. In this guide, we’ explain the formula, walk you through each step you need to take to integrate by parts, and solve example problems so you can become an integration by parts expert yourself.

##
What Is the Integration by Parts Formula?

Integration by parts is

a technique used in calculus to find the integral of a product of functions in terms of the integral of their derivative and antiderivative.

Basically, if you have an equation with the antiderivative two functions multiplied together, and you don’t know how to find the antiderivative, the integration by parts formula transforms the antiderivative of the functions into a different form so that it’s easier to find the simplify/solve.

Here is the formula:

∫ f(x)g’(x) dx = f(x)g(x) − ∫ f’(x)g(x) dx

You start with the left side of the equation (the antiderivative of the product of two functions) and transform it to the right side of the equation.

The integration by parts formula can also be written more compactly,

with u substituted for f(x), v substituted for g(x), dv substituted for g’(x) and du substituted for f’(x):

∫ u dv = uv − ∫ v du

You can use integration by parts when you have to find the antiderivative of a complicated function that is difficult to solve without breaking it down into two functions multiplied together. It may not seem like an incredibly useful formula at first, since neither side of the equation is significantly more simplified than the other, but as we work through examples, you’ll see how useful the integration by parts formula can be for solving antiderivatives.

##
How to Solve Problems Using Integration by Parts

There are five steps to solving a problem using the integration by parts formula:

#1: Choose your u and v

#2: Differentiate u to Find du

#3: Integrate v to find ∫v dx

#4: Plug these values into the integration by parts equation

#5: Simplify and solve

It may seem complicated to integrate by parts, but using the formula is actually pretty straightforward. The first three steps all have to do with choosing/finding the different variables so that they can be plugged into the equation in step four.

You’ll have to have a solid grasp of how to differentiate and integrate, but if you do, those steps are easy.

In general, your goal is for

du

to be simpler than

u

and for the antiderivative of

dv

to not be any more complicated than

v

. Basically, you want the right side of the equation to stay as simple as possible to make it easier for you to simplify and solve. However, don’t stress too much over choosing your u and v. If your first choices don’t work, just switch them and integrate by parts with your new u and v to see if that works better.

Once you have your variables, all you have to do is simplify until you no longer have any antiderivatives,

and you’ve got your answer! Keep reading to see how we use these steps to solve actual sample problems.

##
Integration by Parts Examples

Here are three sample problems of varying difficulty. Try to solve each one yourself, then look to see how we used integration by parts to get the correct answer.

###
Example #1: Find ∫ xsin(x) dx

If you were to just look at this problem, you might have no idea how to go about taking the antiderivative of xsin(x). This is where integration by parts comes in! The first step is to select your

u

and

dv

. With “x” as

u

, it’s easy to get

du

, so let’s start there.

u=

x

dv=

sin(x)

For steps 2 and 3, we’ll differentiate

u

and integrate

dv

to get

du

and

v

. The derivative of x is dx (easy!) and the antiderivative of sin(x) is -cos(x).

du=

dx

v=

-cos(x)

Now it’s time to plug those variables into the integration by parts formula: ∫ u dv = uv − ∫ v du. This gives us:

∫ xsin(x) dx = x(-cos(x)) – ∫ -cos(x) dx

Next, work the right side of the equation out to simplify it. First distribute the negatives:

= -xcos(x) + ∫ cos(x) dx

The antiderivative of cos(x) is sin(x), and don’t forget to add the arbitrary constant, C, at the end:

=

-xcos(x) + sin(x) + C

That’s it, you found the antiderivative!

###
Example #2: Find ∫ x

^{
2
}

ln(x) dx

Again, first we’ll choose a

u

and a

dv

.

u=

ln(x)

dv=

x

^{
2
}

Then we’ll use that information to determine

du

and

v

. The derivative of ln(x) is (1/x) dx, and the antiderivative of x

^{
2
}

is (⅓)x

^{
3
}

.

du=

(1/x) dx

v=

(⅓)x

^{
3
}

Now that we have all the variables, let’s plug them into the integration by parts equation:

∫x

^{
2
}

ln(x)dx=ln(x)⋅(⅓)x

^{
3
}

−∫(⅓)x

^{
3
}

⋅(1/x) dx

All that’s left now is to simplify! First multiply everything out:

= (x

^{
3
}

ln(x))/3 – ∫x

^{
2
}

/3 dx

Then take the antiderivative of ∫x

^{
2
}

/3. Add the constant, and you’re done; there are no more antiderivatives left in the equation:

=

(x

^{
3
}

ln(x))/3 – (1/9)x

^{
3
}

+ C

###
Example #3: Find ∫ ex sin(x) dx

Again, choose a

u

and a

dv

:

u=

sin(x)

dv=

ex

Find

du

and

v

(the derivative of sin(x) is cox(x) and the antiderivative of ex is still just ex.

du=

cos(x)

v=

ex

Enter those variables into the equation:

∫ex sin(x) dx = sin(x) ex -∫cos(x) ex dx

Things are still pretty messy, and the “∫cos(x) ex dx” part of the equation still has two functions multiplied together.

Sometimes, when you use the integrate by parts formula and things look just as complicated as they did before, with two functions multiplied together, it can help to use integration by parts again.

Let’s try it.

Focusing just on the “∫cos(x) ex dx” part of the equation, choose another

u

and

dv

. The derivative of cos(x) is -sin(x), and the antiderivative of ex is still ex (at least that’s easy!).

u=

cos(x)

dv=

ex

du=

-sin(x)

v=

ex

Plug these new variables into the formula again:

∫ex sin(x) dx = sin(x) ex – (cos(x) ex −∫−sin(x) ex dx)

Now simplify:

∫ex sin(x) dx = ex sin(x) – ex cos(x) −∫ ex sin(x)dx

We can move the “−∫ ex sin(x)dx” from the right side of the equation over to the left:

2∫ex sin(x) dx = ex sin(x) − ex cos(x)

Simplify this again, and add the constant:

∫ex sin(x) dx = ex (sin(x) – cos(x)) / 2 + C

There are no more antiderivatives on the right side of the equation, so there’s your answer! We were able to find the antiderivative of that messy equation by working through the integration by parts formula twice.

##
Summary: How to Integrate by Parts

The integration by parts formula can be a great way to find the antiderivative of the product of two functions you otherwise wouldn’t know how to take the antiderivative of. You’ll need to have a solid knowledge of derivatives and antiderivatives to be able to use it, but it’s a straightforward formula that can help you solve various math problems. The steps are:

#1: Choose your u and v

#2: Differentiate u to Find du

#3: Integrate v to find ∫v dx

#4: Plug these values into the integration by parts equation

#5: Simplify and solve

##
What’s Next?

Wondering which math classes you should be taking?

Learn which math classes high schoolers should take by reading our guide.

Interested in

math competitions

like the

International Math Olympiad

? Read our guide to learn how to pass the qualifying tests.

Struggling with the math section of the SAT or ACT?

We have complete guides to

SAT Math

and

ACT Math

to teach you everything you need to ace these sections.