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4.6
Slide 4.6 - 1
Integration Techniques:
Integration by Parts
OBJECTIVES
Evaluate integrals using the formula
for integration by parts.
Solve applied problems involving
integration by parts.
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4.6 Integration Techniques:
Integration by Parts
THEOREM 7
The Integration-by-Parts Formula
∫ u dv = uv − ∫ v du
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Slide 4.6 - 3
1
4.6 Integration Techniques:
Integration by Parts
Tips on Using Integration by Parts:
1. If you have had no success using substitution, try
integration by parts.
2. Use integration by parts when an integral is of the
form
∫ f (x)g(x) dx.
Match it with an integral of the form
∫ u dv
by choosing a function to be u = f (x), where f (x)
can be differentiated, and the remaining factor to be
dv = g(x) dx, where g(x) can be integrated.
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Slide 4.6 - 4
4.6 Integration Techniques:
Integration by Parts
3. Find du by differentiating and v by integrating.
4. If the resulting integral is more complicated than the
original, make some other choice for u = f (x) and
dv = g(x) dx.
5. To check your solution, differentiate.
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Slide 4.6 - 5
4.6 Integration Techniques:
Integration by Parts
Example 1: Evaluate: ∫ ln x dx.
Let u = ln x and dv = dx.
Then,
du =
1
dx and v = x.
x
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Slide 4.6 - 6
2
4.6 Integration Techniques:
Integration by Parts
Example 1 (concluded):
Then, the Integration-by-Parts Formula gives
∫ ln x dx
 1
= (ln x)x − ∫ x   dx
 x
=
x ln x − ∫ dx
=
x ln x − x + C
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Slide 4.6 - 7
4.6 Integration Techniques:
Integration by Parts
Example 2: Evaluate:
∫ x ln x dx.
Let’s examine several choices for u and dv.
Attempt 1: Let u = 1 and dv = x ln x dx.
This will not work because we do not know how to
integrate dv = x ln x dx.
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Slide 4.6 - 8
4.6 Integration Techniques:
Integration by Parts
Example 2 (continued):
u = x ln x and dv = dx.
Attempt 2: Let
 1

Then du =  x ⋅ + 1(ln x ) dx and v = x.
 x

Using the Integration-by-Parts Formula, we have
∫ (x ln x)dx
= (x ln x)x − ∫ x ((1 + ln x)dx )
=
x 2 ln x − ∫ (x + x ln x)dx
This integral seems more complicated than the
original.
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Slide 4.6 - 9
3
4.6 Integration Techniques:
Integration by Parts
Example 2 (continued):
u = ln x and dv = x dx.
Attempt 3: Let
1
x2
dx and v = .
x
2
Using the Integration-by-Parts Formula, we have
x2
x2  1 
∫ (x ln x)dx = ln x ⋅ 2 − ∫ 2  x  dx
Then du =
x2
1
ln x − ∫ xdx
2
2
x2
x2
ln x − + C
2
4
=
=
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Slide 4.6 - 10
4.6 Integration Techniques:
Integration by Parts
Example 3: Evaluate: ∫ x 5x + 1 dx.
Let
u = x and dv = (5x + 1)1 2 dx.
2
32
Then du = 1⋅ dx and v = (5x + 1) .
15
Using the Integration by Parts Formula gives us
2
2
32
32
∫ x( 5x + 1)dx = x ⋅ 15 (5x + 1) − ∫ 15 (5x + 1) dx
2
2 2
= x ⋅ (5x + 1)3 2 −
⋅ (5x + 1)5 2 + C
15
25 15
2
4
=
x(5x + 1)3 2 −
(5x + 1)5 2 + C
15
375
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Slide 4.6 - 11
4.6 Integration Techniques:
Integration by Parts
2
Example 4: Evaluate: ∫ ln x dx
1
Note that we already found the indefinite integral in
Example 1. Now we evaluate it from 1 to 2.
∫
2
1
ln x dx =
[x ln x − x ]12
= (2 ln 2 − 2) − (1⋅ ln1 − 1)
=
2 ln 2 − 2 − 0 + 1
=
2 ln 2 − 1
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Slide 4.6 - 12
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4.6 Integration Techniques:
Integration by Parts
7
Example 5: Evaluate ∫ x 2 e − x dx to find the area of
0
the shaded region shown below.
Let
u = x 2 and dv = e − x dx.
Then du = 2 x dx and v = −e − x .
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Slide 4.6 - 13
4.6 Integration Techniques:
Integration by Parts
Example 5 (continued):
Using the Integration-by-Parts Formula gives us
∫x
2
(e− x dx) =
=
x 2 (−e− x ) − ∫ −e− x (2x dx)
−x 2 e− x + ∫ 2xe− x dx
To evaluate the integral on the right, we can apply
integration by parts again, as follows.
Let
u = 2x and dv = e− x dx.
Then du = 2 dx and v = −e − x .
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Slide 4.6 - 14
4.6 Integration Techniques:
Integration by Parts
Example 5 (continued):
Using the Integration-by-Parts Formula again gives us
∫ 2x(e
−x
dx) = 2x(−e− x ) − ∫ −e− x (2 dx)
=
−2xe− x − 2e− x + C
Then we can substitute this solution into the formula
on the last slide.
∫x e
2 −x
dx = −x 2 e − x − 2xe− x − 2e− x + C
=
−e− x (x 2 + 2x + 2) + C
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Slide 4.6 - 15
5
4.6 Integration Techniques:
Integration by Parts
Example 5 (concluded):
Then, we can evaluate the definite integral.
∫
7
0
7
x 2 e− x dx =  −e− x (x 2 + 2x + 2) + C  0
=  −e −7 (7 2 + 2 (7 ) + 2)  −  −e−0 (0 2 + 2 (0 ) + 2) 
= −65e−7 + 2
≈ 1.94
Slide 4.6 - 16
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4.6 Integration Techniques:
Integration by Parts
Example 7: Evaluate
∫ x e dx.
3 x
When you try integration by parts on this problem, you
will notice a pattern. Using tabular integration can
greatly simplify your work.
Slide 4.6 - 17
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4.6 Integration Techniques:
Integration by Parts
Example 7 (continued):
f(x) and Repeated
Derivatives
x3
Sign of
Product
(+)
g(x) and Repeated
Integrals
ex
3x2
(–)
ex
6x
(+)
ex
6
(–)
ex
0
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ex
Slide 4.6 - 18
6
4.6 Integration Techniques:
Integration by Parts
Example 7 (concluded):
Add the products along the arrows, making the
alternating sign changes, to obtain
∫ x e dx = x e
3 x
3 x
− 3x 2 e x + 6xe x − 6e x + C
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Slide 4.6 - 19
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