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.
On this page, you should learn about
- integration by parts
- repeated integration by parts
The following videos will help you understand all the concepts from this page
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!
Print from here
This is the formula for integration by parts.
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
Here is a quiz that gets you to practice making the correct choice for u in the formula
START QUIZ!
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 ?
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 ?
You can evaluate \(\int { \sin(x)\cdot \cos ^{ 2 }{ (x) } } dx\) using integration by parts
\(\int { u \cdot \frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
Which part do you use for u ?
Here is a quiz that gets you to practise the method of integration by parts
START QUIZ!
Find the values of a , b and d in the following solution (C is the constant of integration)
\(\int { { x }^{ 3 }\cdot ln(x) } dx\)
= \(\frac { { x }^{ 4 } }{ a } ln(x)-\int { \frac { { x }^{ b} }{ 4 } } dx\)
= \(\frac { { x }^{ 4 } }{ a } ln(x)-\frac { { x }^{ 4 } }{ d } + C\)
a =
b =
d =
You need to use integration by parts:
u = ln(x) \(\frac {dv}{dx}=x^3\)
\(\frac {dv}{dx}=\frac {1}{x}\) v = \(\frac {x^4}{4}\)
Drop the correct answers inside the boxes to complete this integral
+ C x2 lnx ex x
\(\int {x\cdot { e }^{ x }} dx\) = ex - ex
You need to use integration by parts:
u = x \(\frac {dv}{dx}=e^x\)
\(\frac {dv}{dx}=1\) v = \(e^x\)
Find the values of a , b , d and e in the following solution (C is the constant of integration)
\(\int { { x }^{ 2 }\cdot sin(2x) } dx\)
= \(-\frac { { x }^{ 2 } }{ 2 } cos(2x)\quad +\quad \int { { x } \cdot \ a } \ dx\)
= \(-\frac { { x }^{ 2 } }{ 2 } cos(2x)\quad +\quad \frac { { x }^{ 2 } }{ b } \cdot d\quad -\quad \int { \frac { sin(2x) }{ 2 } } dx\)
= \(-\frac { { x }^{ 2 } }{ 2 } cos(2x)\quad +\quad \frac { { x }^{ 2 } }{ b } \cdot d\quad +\quad \frac { cos(2x) }{ e } \quad +\quad C\)
a =
b =
d =
e =
In this question you have to do integration by parts 2 times!
Here is a solution to a very similar question.
Evaluate \(\int _{ 1 }^{ e }{ ln(x) dx } \)
This is a classic integral that you should try to remember. It uses integration by parts
\(\int { u \cdot\frac { dv }{ dx } dx } \quad =\quad uv\quad -\quad \int { v\cdot \frac { du }{ dx } } dx\)
| u = lnx | \(\frac{dv}{dx}=1\) |
| \(\frac{du}{dx}=\frac{1}{x}\) | v = x |
\(\int _{ 1 }^{ e }{ ln(x)dx } ={ \left[ x\cdot ln(x)\ -\ x \right] }_{ 1 }^{ e }\)
= \(\left[ e\cdot ln(e)\ -\ e \right] \ -\ \left[ 1\cdot ln(1)\ -\ 1 \right] \)
= \(\left[ e\cdot 1\ -\ e \right] \ -\ \left[ 1\cdot 0\ -\ 1 \right] \)
= 1
Video Solution
How much of Integration by Parts have you understood?
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