Integration by parts is a method of integration that we use to integrate the product (usually !) of two functions. The aim is to change this product into another one that is easier to integrate. Although the formula looks quite odd at first glance, the technique is not too complicated.
\(\int { u \cdot\frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
The tricky part is
1) deciding when we use it (the question will never say ‘solve this integral by the method of integration by parts’) and
2) deciding which part of the question we call u and which part \(\frac { dv }{ dx }\).
If you work through the resources on this page, you should be able to master these two things.
Key Concepts
On this page, you should learn about
integration by parts
repeated integration by parts
Essentials
The following videos will help you understand all the concepts from this page
What and Why
The formula for integration by parts is found from the Product Rule for differentiation. Here's how
Integration by parts is used for integrating the product of two functions (there are one or two exceptions to this that we will see later). We use this technique when Integration by Substitution (or recognition) does not work. You can see in the table below some examples of when we use each technique
Integration by Substitution
Integration by Parts
\(\int { 2x\cdot { e }^{ -3{ x }^{ 2 } }dx } \)
\(\int { 2x\cdot { e }^{ -3{ x } }dx } \)
\(\int { sinx\cdot { e }^{ cosx }dx } \)
\(\int { sinx\cdot { e }^{ x }dx } \)
\(\int { x\cdot \sqrt { x+1 } dx } \)
\(\int { x\cdot \sqrt { x+1 } dx } \)
Notice that it is sometimes possible to use either technique, as is the case in the last example!
Deciding on u and dv/dx 1
When using the formula for integration by parts you need to learn which part of the product you are going to call u and which part \(\frac { dv }{ dx }\).
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
Generally, you should
a) Let u be the function that gets simpler when you differentiate
b) Let \(\frac { dv }{ dx }\)be the function that doesn't get too complicated when you integrate
Here is a video that will help you with that. It shows what happens when you make the correct choice and what happens when you make the wrong choice.
Use integration by parts to find \(\int { 2x\cdot { e }^{ -3{ x }}dx } \)
Deciding on u and dv/dx 2
There following list might help you with your choice. Look at the two functions in your question. The one that appears first on the following list is the one you should choose to be u
Logs
Inverse Trig
Algebra
Trig / Exponential
Here's another example
Use integration by parts to find \(\int { x^ 2\cdot lnxdx } \)
Repeating the Process 1
Sometimes you have to apply the method of integration by parts more than once!
Use integration by parts to find \(\int { x^{ 2 }\cdot sin\left( \frac { x }{ 2 } \right) dx } \)
Repeating the Process 2
Sometimes integration by parts really doesn't seem to help and we end up going round in circles! Here's an example where that is the case and a clever solution to it!
Use integration by parts to find \(\int { e^{ x }\cdot cos2xdx } \)
Special Integrals
Integration by parts is mainly used for integrating the product of two functions. However, we can use it to find the integral of ln(x) and arctrigonometric functions too. The second question requires integration by substitution as well as integration by parts!
Use integration by parts to find \(\int { lnx } dx\) and \(\int { arcsinx } dx\)
You can evaluate \(\int { { x\cdot e }^{ x } } dx\)using integration by parts
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
Which part do you use for u ?
u should be the first type of function that you see in the list below
1) Logarithmic
2) Inverse Trigonometric
3) Algebraic
4) Trigonometric/Exponential
You can evaluate \(\int { x^3 \cdot ln(x) } dx\)using integration by parts
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
Which part do you use for u ?
u should be the first type of function that you see in the list below
1) Logarithmic
2) Inverse Trigonometric
3) Algebraic
4) Trigonometric/Exponential
You can evaluate \(\int { 1\cdot arctan(x) } dx\) using integration by parts
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
Which part do you use for u ?
This question is an interesting one since it doesn't look like a product of two parts. Here are a couple of solutions to a question like this one
You can evaluate \(\int { 2x\cdot arctan(x) } dx\)using integration by parts
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
Which part do you use for u ?
u should be the first type of function that you see in the list below
1) Logarithmic
2) Inverse Trigonometric
3) Algebraic
4) Trigonometric/Exponential
You can evaluate \(\int { { sinx\cdot e }^{ x } } dx\)using integration by parts
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
Which part do you use for u ?
In this question it doesn't matter, but you have to do integration by parts twice. The question seems to go around in circles. Here's a solution to a question like this
You can evaluate \(\int { \sin(x)\cdot \cos ^{ 2 }{ (x) } } dx\)using integration by parts
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
This question requires you to use Integration by Parts
In the second part you will need to integrate a rational function. Since the degree of the numerator is equal to the degree of the denominator, you need to divide the polynomials