7.2. Integration by Parts

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7.2 Integration by Parts 1. Let u = 3x and dv = ex dx. Then du = 3 dx and v = ex .

Z

x

3xe dx = uv −

Z

v du

= 3xex − 3

Z

ex dx

= 3xex − 3ex +C 2. Let u = x2 and dv = e−x dx. Then du = 2x dx and v = −e−x .

Z

x2 e−x dx = uv −

Z

v du

2 −x

= −x e

Z

+

2xe−x dx

Let u = 2x and dv = e−x dx. Then du = 2 dx and v = −e−x .

Z

x2 e−x dx = −x2 e−x +

Z

2xe−x dx Z

2 −x

+ (uv − v du)   Z 2 −x −x −x = −x e + −2xe − −2xe dx  
= −x2 e−x − 2xe−x + − −2(−e−x ) +C = −x e

= −x2 e−x − 2xe−x − 2e−x +C 3. Let u = ln(3x + 2) and dv = 1 dx. Then du =

Z

3 3x+2 dx

and v = x.

ln(3x + 2)dx = uv −

Z

v du

3 dx 3x + 2 Z 3x = xln|3x + 2|− dx 3x + 2 Z

= xln|3x + 2|−

Let u = 3x + 2 and x = 224

u−2 3 .

Then du = 3 dx.

x

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Chapter 7. Integration Techniques, Solution Key

3x dx Z3x + 2  u−2 = xln|3x + 2|− du 3u Z    1 2 = xln|3x + 2|− − du 3 3u 1 2 = xln|3x + 2|− u + ln|u| 3 3 2 1 = xln|3x + 2|− (3x + 2) + ln|3x + 2| 3 3

Z

Z

ln(3x + 2)dx = xln|3x + 2|−

4. Let u = sin−1 x and dv = 1 dx. Then du = √ 1 2 dx and v = x. 1−x Z

sin−1 xdx = uv −

Z

v du

= x sin−1 x −

Z

1 √ dx 1 − x2

sin−1 xdx = x sin−1 x −

Z

x √ dx 1 − x2

Let u = 1 − x2 . Then du = −2x dx. Z

−1 1 u 2 du 2 1 1 u2 −1 = x sin x + 1 +C 2 2 p = x sin−1 x + 1 − x2 +C

= x sin−1 x +

5.

R

Z

sec3 x dx = sec2 x sec x dx R

Let u = sec x and dv = sec2 x dx. Then du = tan x dx and v = tan x. Z

3

Z

sec x dx =

sec2 x sec x dx

= uv −

Z

v du

= (sec x)(tan x) − = (sec x)(tan x) − = (sec x)(tan x) − Z

2

Z

sec3 x dx = (sec x)(tan x) +

Z

(tan2 x) sec x dx

Z

(sec2 x − 1) sec x dx

Z

3

sec x dx +

Z

sec x dx

Z

sec x dx

= (sec x)(tan x) + ln|sec x + tan x| 1 1 sec3 x dx = (sec x)(tan x) + ln|sec x + tan x|+C 2 2 225

7.2. Integration by Parts

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6. Let u = ln(3x) and dv = 2x dx. Then du =

Z

3 3x dx

= 1x dx and v = x2 .

2xln(3x)dx = uv −

Z

v du

x2 dx Z x = x2 ln(3x) − x dx 2

= x ln(3x) −

Z

x2 2 2 x = x2 ln(3x) − +C 2 = x2 ln(3x) −

7. Let u = lnx. Then du = 1x dx.

Z

(lnx)2 dx = x

Z

u2 du

u3 +C 3 (lnx)2 = +C 3

=

8. Let u = 5x − 2. Then du = 5 dx and x =

Z

x

u+2 5 .

Z √ 1 5x − 2dx = 5

1 = 5 =

Z

1 5



u+2 5



√ u du

1

3

u 2 2u 2 + du 5 5 ! 5 3 u2 2u 2
+ 3
5 2 5 52 5

5

2(5x − 2) 2 2(5x − 2) 2 = + +C 125 75 Let u = x and dv =

√ 5x − 2dx.Then du = dx and v =

3

1 (5x−2) 2 3 5 2

Z

x

9. Let u = x2 and dv = e5x dx.

226

=

3 2 2 15 (5z − 2) .

Z √ 3 3 2 2 5x − 2dx = x(5x − 2) 2 − (5x − 2) 2 dx 15 15 5 3 2 2 1 (5x − 2) 2 2 = x(5x − 2) − × 5 15 15 5 2 3 5 2 4 = x(5x − 2) 2 − (5x − 2) 2 +C 15 375

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Chapter 7. Integration Techniques, Solution Key

TABLE 7.1: Alternate signs + − + −

u and its derivatives x2 2x 2 0

Z

dv and its antiderivatives e5x 1 5x 5e 1 5x 25 e 1 5x 125 e

1 2 2 5x x2 e5x dx = x2 e5x − xe5x + e +C 5 25 125

10. Let u = x2 and dv = ex dx.

TABLE 7.2: Alternate signs + − + −

u and its derivatives x2 2x 2 0

Z1

dv and its antiderivatives ex ex ex ex

x2 ex dx = (x2 ex − 2xex + 2ex )|10

0

= e − 2e + 2e − (2) = e−2 11. Let u = ln(x + 1) and dv = 1 dx. Then du =

1 x+1 dx

Z3

ln(x + 1)dx =

and v = x.

xln(x + 1)|31 −

1

Z3

x dx x+1

1

=

xln(x + 1)|31 −

Z3 

 1 1− dx x+1

1

= [xln(x + 1) − x + ln(x + 1)]31 = 3ln4 − 3 + ln4 − (ln2 − 1 + ln2) = 4ln4 − 2 − 2ln2 = 8ln2 − 2ln2 − 2 = 6ln2 − 2

227