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17calculus > integrals > integration by parts

Topics You Need To Understand For This Page

Calculus Main Topics


Related Topics and Links

related topics on other pages

integration by substitution

external links you may find helpful

integration by parts youtube playlist

WikiBooks: Integration By Parts

ATTENTION INSTRUCTORS: The new 2018 version of 17calculus will include changes to the practice problem numbering system. If you would like advance information to help you prepare for spring semester, send us an email at 2018info at 17calculus.com.

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Integration by Parts

Be careful that you know the difference between integration by substitution and integration by parts.

Integration by parts is an extremely important and useful integration technique. Most integrals you come across won't be in a simple form. The idea is to 'reduce' or alter the original integral by breaking it up into pieces that can then be evaluated using the techniques you know so far. You will find that even a small change in an integral will often make a big change in the technique used and the final solution.

Integration by parts refers to the use of the equation \(\int{ u~dv } = uv - \int{ v~du }\).
It is important to read the next section to understand where this comes from.

Deriving the Integration By Parts Equation

The integration by parts equation comes from the product rule for derivatives. We will show an informal proof here.
Let u and v be functions of t.

Using the product and chain rules, we take the derivative of \(u \cdot v\) to get

\(\displaystyle{ \frac{d}{dt}[ u \cdot v] = u \frac{dv}{dt} + v \frac{du}{dt} }\)

Now we isolate the \(\displaystyle{ u \frac{dv}{dt} }\) term to get

\(\displaystyle{ u \frac{dv}{dt} = \frac{d}{dt}[ u \cdot v] - v \frac{du}{dt} }\)

Next, we integrate both sides with respect to \(t\) to get

\(\displaystyle{ \int{ u \frac{dv}{dt} ~dt } = }\) \(\displaystyle{ \int{ \frac{d}{dt}[ u \cdot v] ~dt } - \int{ v \frac{du}{dt} ~dt } }\)

Using the idea of differentials, we end up with

\(\displaystyle{ \int{ u ~dv } = u \cdot v - \int{ v ~du } }\)

Here is a video showing the same derivation as above.

PatrickJMT - Deriving the Integration by Parts Formula - Easy!

Using Integration By Parts

The key to using integration by parts is learning what to choose for u and dv from the given integral. This is something you will develop a feel for as you get experience working these problems. However, there are some guidelines you can use as you are learning.

General Guidelines
1. Choose the dv term to make integration easy to get v since integration is usually more difficult than taking the derivative.
2. If you have a natural log term, always choose it to be equal to u. Remember, \((\ln x)'=1/x\) but \(\int{\ln(x)~dx} \neq 1/x \). [ one of the practice problems solves the integral \(\int{\ln(x)~dx}\) demonstrating what happens when you don't do this ]
3. If you have a polynomial, set it equal to u since taking the derivative to get du will reduce the power by one. [ the reduction formula video below shows why this works ]

Specific Guidelines
There is a rule of thumb called the LIATE rule that suggests which function to choose for u. By going down the list, try choosing the first function you come across as u. This rule does not work ALL the time but it is a good place to start as you are learning integration by parts.



logarithmic functions like \(\ln(x)\)


inverse trig functions like \(\arcsin(x)\)


algebraic functions like polynomials


trigonometric functions like \(\sin(x)\)


exponential functions like \(e^{3x}\)

source: wikipedia

Other Situations That You Will See
- You will come across an integral where you may need to use integration by parts multiple times. This is especially true when you have a polynomial. [ the reduction formula video below demonstrates this ]
- Another case you will run into is that, after using integration by parts once or more, you may end up with the same integral you started out with. It looks like you don't accomplish anything or it loops back on itself. However, there is a way to solve these types of integrals. [ one of the practice problems solves the integral \(\int{e^{2x}\sin(x)~dx}\) showing this ]

Reduction Formulas

Here are three videos showing how to derive reduction formulas. Although you will may never use the results of these videos directly, they are very good videos to watch to help you understand integration by parts.

In the first video, he derives the reduction formula for \( \int{ x^n e^x ~dx} \) and, in the process, demonstrates two of the main points discussed above, (1) choosing u as the polynomial reduces the power in the next step, and (2) doing integration by parts multiple times is sometimes necessary.

Dr Chris Tisdell - Reduction formula: integration

In this video, she derives the reduction formula for \( \int{ (\ln x)^n ~dx} \).

Krista King Math - Integration by parts to prove the reduction formula

In this video, he derives the reduction formula for \( \int{ \sec^n(x) ~dx} \).
Although the notation gets a bit messy, this is a good video demonstrating the use of integration by parts. If you get lost, just try to get the main points he is making.

PatrickJMT - Reduction Formula for: Integral of [sec(x)]^n dx

Okay, time for you to try some practice problems.

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Practice Problems

Instructions - - Unless otherwise instructed, use integration by parts to evaluate these integrals, giving your answers in exact form.

Level A - Basic

Practice A01




Practice A02




Practice A03



Practice A04




Practice A05



Practice A06



Practice A07



Practice A08



Practice A09



Practice A10



Practice A11



Practice A12



Level B - Intermediate

Practice B01




Practice B02




Practice B03

\(\displaystyle{\int{(\ln x)^2dx}}\)


Practice B04



Practice B05



Practice B06



Practice B07



Practice B08



Practice B09



Practice B10




Level C - Advanced

Practice C01

\(\displaystyle{\int{(4t^2-8t+5)~e^{t^2-3t}dt}}\)     Tricky! - Hint



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