Integration By Parts
R
As a motivating example let’s consider xex dx. At first glance, integration appears difficult, as none of our
previous methods apply. In order to gain some intuition we investigate the derivative of xex . The Product
Rule gives
d x
d x
xe = ex + xex =⇒ xex =
xe − ex .
dx
dx
Recalling that antidifferentiation undoes a derivative, we find
Z
Z
Z
d x
xe dx − ex dx
xex dx =
dx
= xex − ex + C.
We can view this from a more general perspective
Z
Z
Z
d
d
0
0
0
(f (x)g(x)) = f (x)g (x) + g(x)f (x) =⇒
(f (x)g(x)) dx = f (x)g (x) dx + g(x)f 0 (x) dx.
dx
dx
Recalling that integration “undoes” differentiation and reorganizing terms leads to
Z
Z
f (x)g 0 (x) dx = f (x)g(x) − g(x)f 0 (x) dx
(1.1)
which is know as Integration by Parts (IBP). It is often more convenient to let u = f (x) and v = g(x).
Then du = f 0 (x)dx and dv = g 0 (x)dx so that Equation (1.1) becomes
Z
Z
u dv = uv − v du.
Observations:
• Integration by Parts “undoes” the Product Rule of differentiation.
• Integration by Parts is typically applied to integrate a product.
R
• Integration by Parts does not solve
It allows us to express a complicated integral ( f (x)g 0 (x) dx)
R an integral.
in terms of a simpler integral ( g(x)f 0 (x) dx).
Z
Example 1. Evaluate
xex dx.
Intuition: The integrand xex is a product of the form u dv with u = x and dv = ex . We can antidifferentiate
dv to find v = ex and differentiate u to find du = dx. So it appears reasonable to apply IBPs.
u=x
−→ v = ex
(Note: The arrows indicate how we apply IBPs. Multiply uv and then
du
.
dv = ex dx
R = dx
subtract v du.) IBPs gives
Using
Z
x
x
xe dx = xe −
Z
ex dx
= xex − ex + C.
Observations:
1
Calculus II Resources
Integration Techniques
• The choices of u and dv are not typically interchangeable. If u = ex and dv = xdx, then IBPs yields
Z
Z
1 2 x
1 2 x
x
xe dx = x e −
x e dx
2
2
and the resulting integral has become more difficult to evaluate.
• We choose u and dv in a way that dv is something we can antidifferentiate, and ideally du is ‘simpler’
than u.
• IBPs can be applied to integrate functions of the form p(x)ekx with p(x) a polynomial and k a constant
by letting u = p(x) and dv = ekx .
See solution video
Z
Example 2. Evaluate
arctan(x) dx.
Intuition: At first glance, the integrand arctan(x) does not appear to be a product of functions. However,
arctan(x) = arctan(x) · 1
and arctan(x) is a function we can easily differentiate.
Using
u = arctan(x)
1
du = 1+x
2 dx
−→
.
v=x
IBPs gives
dv = 1dx
Z
Z
x
dx
1 + x2
Z
1
1
= x arctan(x) −
dw
2
w
1
= x arctan(x) − ln |w| + C
2
1
= x arctan(x) − ln |1 + x2 | + C
2
arctan(x) dx = x arctan(x) −
where in the second step we have used the substitution w = 1 + x2 (dw = 2xdx so that 21 dw = xdx).
Observations:
• IBPs can also be applied to integrate other ‘inverse’ functions. (i.e. arcsin(x), arccos(x), ln(x), . . .)
• Integrals may require multiple techniques of integration.
See solution video
2
Z
xe6x dx.
Example 3. Evaluate
0
Observation: When applying integration by parts to a definite integral, we simply need to add in the limits
of integration. In general,
Z
a
Using
u=x
du = dx
−→
.
b
b Z
f (x)g 0 (x) dx = f (x)g(x) −
a
v = 61 e6x
IBPs gives
dv = e6x dx
2
a
b
g(x)f 0 (x) dx.
Calculus II Resources
Integration Techniques
Z
2 Z 2
1 6x
x 6x e −
e dx
6
6
0
0
2
1 12
1 6x = e −0− e 3
36
0
1 12
1 12
1
= e −
e −
3
36
36
11 12
1
=
e + .
36
36
2
xe6x dx =
0
See solution video
Z
Example 4. Evaluate
x3 ex/3 dx.
Observation: Instead of applying IBPs three times (individually) we will take care of them all at once. In
order to do so, we make a table of u and dv values and combine them appropriately. We differentiate down
the u column until we get to zero. In the dv column we integrate. In the sign column we alternate between
‘+’ and ‘−’.
sign
+
+
STOP
u
x3
3x2
6x
6
0
·
·
·
·
dv = ex/3
3ex/3
9ex/3
27ex/3
81ex/3
We multiply together the entries in each row and put the sign term in front of each product. In this case,
Z
x3 ex/3 dx = (x3 ) 3ex/3 − (3x2 ) 9ex/3 + (6x) 27ex/3 − (6) 81ex/3 .
See solution video
Summary:
• Integration by Parts is an integration technique used to integrate functions of the form
Letting u = f (x) and dv = g 0 (x)dx gives
Z
R
f (x)g 0 (x) dx.
Z
u dv = uv −
v du.
• We choose u and dv in a way that dv is something we can antidifferentiate and ‘ideally’ du is simpler
than u.
• When applying IBPs to a definite integral we simply add in the limits of integration.
2 x
x
• IBPs is well-suited to integrate functions of the form x2 cos(x),
R x e , e sin(x), arcsin(x), ln(x), and
other inverse functions. Depending upon the function f (x), xf (x) dx may be well-suited for IBPs.
See IBPs overview video
Practice Problems: Evaluate the following Integrals
3
Calculus II Resources
Integration Techniques
Z
(1)
Z
(2)
Z
ln(x)
dx
x2
(4)
Z
x
e sin(x) dx
Z
(3)
x ln(x + 1) dx
(5)
1
Z
ln(x + 1) dx
(6)
0
See Solutions
4
x4 ex dx
x sec2 (x) dx