Calculus Diff Chapter 7 Test Review Answers

1. Let

€

u = 2xdu = 2 dx

giving

€

12

11+ u2∫ du = 1

2arctan 2x( )+C

2. Let

€

u =1+ cos xdu = −sin x dx

giving

€

− u2∫ du

€

= − 13 1+ cos x( )3 +C

3. Let

€

u = ln xdu = 1

x dx giving

€

1u2∫ du

€

=−1ln x

+C

4. Let

€

u = −x2

du = −2x du giving

€

−12

eu−4

−16∫ du =

−12eu

−16−4

=−12e−16 − −1

2e−4 =

−12e16

+12e4

5. Let

€

u = x − 3du = dx

giving

€

1udu

0

3∫ =

u12

12

3

0= 2 3 − 2 0 = 2 3

6. Let

€

u = 2xdu = 2 dx

giving

€

12

cosu0

π3∫ du =

12

sinu

π3

0=

12

sin π3( ) − 1

2sin 0( ) =

12

32

%

& '

(

) * =

34

7. Use

€

u = 9x w = 15 sin 5x( )

du = 9 dx dw = cos 5x( ) dx to get 9x( ) 1

5 sin 5x( )( )− 15 sin 5x( ) ⋅9 dx∫

€

= 9x5 sin 5x( ) + 9

25cos 5x( ) +C

8. Use

€

u = sin−1 x w = xdu = 1

1−x2 dx dw = dx to get

€

sin−1 x( ) dx∫

€

= sin−1 x( ) x( ) − x1−x2

dx∫ , use

€

u =1− x2

du = −2x dx to get

€

sin−1 x( ) x( ) − −12( ) 1

u du∫

€

= x sin−1 x + 12 ⋅u12

12

+C

€

= x sin−1 x + 1− x2 +C

9. Use partial fractions,

€

9x + 2x + 6( ) x −1( )

=A

x + 6+

Bx −1

giving

€

9x + 2 = A x −1( ) + B x + 6( ), the two

equations are

€

9 = A + B2 = −A + 6B

, solving gives

€

A =527

and

€

B =117

. The integral is now

€

527

x + 6+

117

x −1∫ dx

€

=527ln x + 6 +

117ln x −1 +C

10. Use partial fractions,

€

x2 − x + 4x x − 5( ) x + 4( )

=Ax

+B

x − 5+

Cx + 4

giving

€

x2 − x + 4 = A x − 5( ) x + 4( ) + Bx x + 4( ) +Cx x − 5( ) or

x2 − x + 4 = Ax2 − Ax − 20A+Bx2 + 4Bx +Cx2 − 5Cx , the three equations are 1= A+B+C

−1= −A+ 4B− 5C4 = −20A

,

solving last equation gives

€

A = −15 , then have

1= − 15 +B+C

−1= − − 15( )+ 4B− 5Cwhich  gives

€

B = 815 and

€

C = 23 . The integral is now

€

−15x

+8

15x − 5

+23

x + 4∫ dx

€

=−15ln x +

815ln x − 5 +

23ln x + 4 +C

11. Use partial fractions with unfactorable quadratic,

€

x2 + 2x + 3x x2 +1( )

=Ax

+Bx +Cx2 +1

giving

€

x2 + 2x + 3 = A x2 +1( ) + Bx +C( )x , the three equations are

€

1 = A + B2 = C3 = A

giving

€

B = −2 . The integral

is now

€

3x

+−2x + 2x2 +1

dx∫

€

=3x

+−2xx2 +1

+2

x2 +1 dx∫ , in the second integral use

€

u = x2 +1du = 2x dx

giving 3 1x

dx∫ −1u

du∫ + 2 1x2 +1

dx∫

€

= 3ln x − lnu + 2tan−1 x +C

€

= 3ln x − ln x2 +1 + 2tan−1 x +C

12. Use partial fractions with a repeated factor,

€

x2 + x − 7x2 x − 3( )

=Ax

+Bx2

+Cx − 3

giving

€

x2 + x − 7 = Ax x − 3( ) + B x − 3( ) +Cx2, the three equations are

€

1 = A +C1 = −3A + B−7 = −3B

, solving gives

€

A = 49 ,

€

B = 73 and

€

C = 59 . The integral is now

€

49x

+73x2 +

59

x − 3∫ dx

€

=49ln x − 7

3x−1 +

59ln x − 3 +C

13. Complete the square to get 3

x2 −10x + −5( )2( )+ 29− −5( )2 dx∫

€

=3

x − 5( )2 + 4 dx∫

€

=34

1x−5( )2

4 +1 dx∫ , let

€

u = x−52

du = 12 dx

giving

€

34⋅ 2 1

u2 +1 du∫

€

= 32 tan

−1 u +C

€

= 32 tan

−1 x−52( ) +C

14. Complete the square to get 5x +3x2 + 6x +32( )+10−32

dx∫

€

=5x + 3x + 3( )2 +1

dx∫ , write as two

integrals 5xx +3( )2 +1

dx∫ +3

x +3( )2 +1 dx∫ =

5 x +3( )x +3( )2 +1

dx∫ +3− 5( ) 3( )x +3( )2 +1

dx∫ , on the first

use

€

u = x + 3( )2 +1du = 2 x + 3( ) dx

and on the second use

€

w = x + 3dw = dx

giving

€

52

1u

du∫ −12 1w2 +1

dw∫

€

=52lnu −12tan−1w +C

€

=52ln x + 3( )2 +1 −12tan−1 x + 3( ) +C

15. Let

€

x = 3sinθ and

€

dx = 3cosθ dθ , also

€

9 − x2 = 3cosθ . Then

€

1

x2 9 − x2dx∫

€

=1

3sinθ( )2 3cosθ( )3cosθ dθ∫

€

=19

1sin2θ

dθ∫

€

=19

csc2θ dθ∫

€

=−19cotθ + C

€

=− 9 − x2

9x+ C

16. Let

€

x = tanθ and

€

dx = sec2θ dθ , also

€

1+ x2 = secθ . Then 1

1+ x2( )32dx∫ = 1

secθ( )3sec2θ dθ∫ = 1

secθ dθ∫ = cosθ∫  dθ

= sinθ +C = x

1+ x2+C

17. x

x2 − 5 3

∞∫ dx = lim

b→∞ x

x2 − 5 3

b∫ dx , let

€

u = x2 − 5du = 2x dx

giving limb→∞

12

1u 3

b∫ du

= limb→∞

  12⋅u

12

12 x=3

x=b

= limb→∞

  x2 − 53

b = ∞2 − 5 − 32 − 5 =∞− 2 =∞ , diverges.

18.

€

2x2 +1

dx1

∞∫ = lim

b→∞  2 1

x2 +1 dx

1

b∫ = lim

b→∞  2 tan−1 x

1

b

€

= 2 tan−1 ∞( ) − tan−1 1( )( )

€

= 2 π2−π4

$

% &

'

( ) = 2

π4$

% & '

( ) =

π2

, converges.

19. 1x 2− x( )

dx3

4∫ = lim

b→4  1

x 2− x( ) dx

3

b∫ , let

u = 2− x

du = − 12 x

−12 dx

giving

limb→4

  − 2 1u

du 3

b∫ = lim

b→4 − 2 ln u

x=3x=b = lim

b→4 − 2 ln 2− x

3

b = −2 ln 2− 4 + 2 ln 2− 3

= −2 ln 0( )+ 2 ln 2− 3( ) = −2 −∞( )+ 2 ln 2− 3( ) =∞+ 2 ln 2− 3( ) =∞ , diverges

20.

€

sin x

cos2 x 0

π2∫ dx = limb→π

2

  sin xcos2 x

dx 0

b∫ , let

€

u = cos xdu = −sin x dx

giving limb→π

2

  − 1u2 du

0

b∫

= limb→π

2

  −u−1

−1x=0

x=b

= limb→π

2

  1u x=0

x=b = lim

b→π2

  1cos x 0

b

€

=1

cos π2( )−

1cos 0( )

€

=10−11

€

= ∞ −1 = ∞ , diverges

21.

€

11+ x( ) x

dx1

∞

∫ =1

x + x3

2 dx

1

∞

∫ . Since

€

x12 ≥ 0 then

€

x12 + x

32 ≥ x

32 giving

€

1

x12 + x

32≤1

x32

.

€

1

x3

2 dx

1

∞

∫ is a p function with

€

p = 32 which is greater than 1, therefore

€

1

x3

2 dx

1

∞

∫ converges.

Since

€

1

x3

2 dx

1

∞

∫ converges and

€

1

x12 + x

32≤1

x32

, then by the direct comparison test

€

11+ x( ) x

dx1

∞

∫ also converges.

22. Since cos x ≤1 then −cos x ≥ −1 then

€

2 − cos x ≥1 then

€

2 − cos x ≥1 and

€

2 − cos xx

≥1x

.

 1x

dx0

1∫ is a p function with p = 1 and therefore diverges. Since

€

2 − cos xx

≥1x

and  1x

dx0

1∫

diverges, then by the direct comparison test

€

2 − cos xx

dx0

1∫ also diverges.