Joe Foster
Integration by Parts
To reverse the chain rule we have the method of u-substitution. To reverse the product rule we also have a method, called
Integration by Parts. The formula is given by:
Theorem (Integration by Parts Formula)
ˆ ˆ
f (x)g(x) dx = F (x)g(x) − F (x)g ′ (x) dx
where F (x) is an anti-derivative of f (x).
Remember, all of the techniques that we talk about are supposed to make integrating easier! Even though this formula
´
expresses one integral in terms of a second integral, the idea is that the second integral, F (x)g ′ (x) dx, is easier to evaluate.
The key to integration by parts is making the right choice for f (x) and g(x). Sometimes we may need to try multiple
options before we can apply the formula. Let’s see it in action.
Example 1 Find
ˆ
x cos(x) dx.
We have to decide what to assign to f (x) and what to assign to g(x). Our goal is to make the integral easier. One thing
to bear in mind is that whichever term we let equal g(x) we need to differentiate - so if differentiating makes a part of the
integrand simpler that’s probably what we want! In this cases differentiating cos(x) gives − sin(x), which is no easier to
deal with. But differentiating x gives 1 which is simpler. So we have,
ˆ ˆ
g(x) = x f (x) = cos(x) x cos(x) dx = x sin(x) − sin(x) dx
g ′ (x) = 1 F (x) = sin(x) = x sin(x) + cos(x) + C
Example 2 Evaluate
ˆ 4
xe−x dx.
0
ˆ 4 4
ˆ 4
xe−x dx = −xe−x − −e−x dx
0 0 0
4
−x −x
−x = −xe − e
g(x) = x f (x) = e 0
′
g (x) = 1 F (x) = −e−x = −4e−4 − e−4 − 0 − e−0
= −5e−4 + 1
= 1 − 5e−4
Page 1 of 8
, MATH 142 - Integration by Parts Joe Foster
Example 3 Evaluate
ˆ
x2 ex dx.
g(x) = x2 f (x) = ex
ˆ ˆ
x2 ex dx = x2 ex − 2 xex dx.
′ x
g (x) = 2x F (x) = e
It’s at this point we see that we still cannot integrate the integral on the write easily. This is okay. Sometimes we may
have to apply the integration by parts formula more than once!
ˆ ˆ
x2 ex dx = x2 ex − 2xex dx
ˆ
2 x x x
g1 (x) = x f1 (x) = ex = x e − 2 xe − e dx
g1′ (x) = 1 F1 (x) = ex
= x2 ex − 2xex + 2ex + C
= x2 − 2x + 2 ex + C
ˆ
The previous technique works for any integral of the form xn emx dx, where n is any positive integer and m is any integer.
What if n was negative? Then this case we would set g(x) = ex .
Example 4 In Example 3 we have to apply the Integration by Parts Formula multiple times. There is a convenient way
to “book-keep” our work. This is done by creating a table. Let’s see how by examining Example 3 again.
ˆ
x2 ex dx.
Let g(x) = x2 and f (x) = ex . Then,
Differentiate g(x) Integrate f (x)
x2 + ex
2x − ex
2 + ex
0 ex
Then the integral is,
ˆ
x2 ex dx = +x2 · ex − 2x · ex + 2 · ex + C
x2 − 2x + 2 ex + C
=
We have actually used the integration by parts formula, but we have just made our lives easier by condensing the work
into a neat table. This method is extremely useful when Integration by Parts needs to be used over and over again.
Page 2 of 8
Integration by Parts
To reverse the chain rule we have the method of u-substitution. To reverse the product rule we also have a method, called
Integration by Parts. The formula is given by:
Theorem (Integration by Parts Formula)
ˆ ˆ
f (x)g(x) dx = F (x)g(x) − F (x)g ′ (x) dx
where F (x) is an anti-derivative of f (x).
Remember, all of the techniques that we talk about are supposed to make integrating easier! Even though this formula
´
expresses one integral in terms of a second integral, the idea is that the second integral, F (x)g ′ (x) dx, is easier to evaluate.
The key to integration by parts is making the right choice for f (x) and g(x). Sometimes we may need to try multiple
options before we can apply the formula. Let’s see it in action.
Example 1 Find
ˆ
x cos(x) dx.
We have to decide what to assign to f (x) and what to assign to g(x). Our goal is to make the integral easier. One thing
to bear in mind is that whichever term we let equal g(x) we need to differentiate - so if differentiating makes a part of the
integrand simpler that’s probably what we want! In this cases differentiating cos(x) gives − sin(x), which is no easier to
deal with. But differentiating x gives 1 which is simpler. So we have,
ˆ ˆ
g(x) = x f (x) = cos(x) x cos(x) dx = x sin(x) − sin(x) dx
g ′ (x) = 1 F (x) = sin(x) = x sin(x) + cos(x) + C
Example 2 Evaluate
ˆ 4
xe−x dx.
0
ˆ 4 4
ˆ 4
xe−x dx = −xe−x − −e−x dx
0 0 0
4
−x −x
−x = −xe − e
g(x) = x f (x) = e 0
′
g (x) = 1 F (x) = −e−x = −4e−4 − e−4 − 0 − e−0
= −5e−4 + 1
= 1 − 5e−4
Page 1 of 8
, MATH 142 - Integration by Parts Joe Foster
Example 3 Evaluate
ˆ
x2 ex dx.
g(x) = x2 f (x) = ex
ˆ ˆ
x2 ex dx = x2 ex − 2 xex dx.
′ x
g (x) = 2x F (x) = e
It’s at this point we see that we still cannot integrate the integral on the write easily. This is okay. Sometimes we may
have to apply the integration by parts formula more than once!
ˆ ˆ
x2 ex dx = x2 ex − 2xex dx
ˆ
2 x x x
g1 (x) = x f1 (x) = ex = x e − 2 xe − e dx
g1′ (x) = 1 F1 (x) = ex
= x2 ex − 2xex + 2ex + C
= x2 − 2x + 2 ex + C
ˆ
The previous technique works for any integral of the form xn emx dx, where n is any positive integer and m is any integer.
What if n was negative? Then this case we would set g(x) = ex .
Example 4 In Example 3 we have to apply the Integration by Parts Formula multiple times. There is a convenient way
to “book-keep” our work. This is done by creating a table. Let’s see how by examining Example 3 again.
ˆ
x2 ex dx.
Let g(x) = x2 and f (x) = ex . Then,
Differentiate g(x) Integrate f (x)
x2 + ex
2x − ex
2 + ex
0 ex
Then the integral is,
ˆ
x2 ex dx = +x2 · ex − 2x · ex + 2 · ex + C
x2 − 2x + 2 ex + C
=
We have actually used the integration by parts formula, but we have just made our lives easier by condensing the work
into a neat table. This method is extremely useful when Integration by Parts needs to be used over and over again.
Page 2 of 8
