Calculus Diff Chapter 7 Test Review Answers 1. Let u = 2 x du = 2 dx giving 1 2 1 1 + u 2 ∫ du = 1 2 arctan 2 x ( ) + C 2. Let u = 1 + cos x du = − sin x dx giving − u 2 ∫ du = − 1 3 1 + cos x ( ) 3 + C 3. Let u = ln x du = 1 x dx giving 1 u 2 ∫ du = −1 ln x + C 4. Let u = − x 2 du = −2 x du giving −1 2 e u −4 −16 ∫ du = −1 2 e u −16 −4 = −1 2 e −16 − −1 2 e −4 = −1 2e 16 + 1 2e 4 5. Let u = x − 3 du = dx giving 1 u du 0 3 ∫ = u 1 2 1 2 3 0 = 2 3 − 2 0 = 2 3 6. Let u = 2 x du = 2 dx giving 1 2 cos u 0 π 3 ∫ du = 1 2 sin u π 3 0 = 1 2 sin π 3 ( ) − 1 2 sin 0 () = 1 2 3 2 % & ' ( ) * = 3 4 7. Use u = 9 x w = 1 5 sin 5 x ( ) du = 9 dx dw = cos 5 x ( ) dx to get 9 x ( ) 1 5 sin 5 x ( ) ( ) − 1 5 sin 5 x ( ) ⋅ 9 dx ∫ = 9 x 5 sin 5 x ( ) + 9 25 cos 5 x ( ) + C 8. Use u = sin −1 x w = x du = 1 1 − x 2 dx dw = dx to get sin −1 x () dx ∫ = sin −1 x ( ) x () − x 1 − x 2 dx ∫ , use u = 1 − x 2 du = −2 x dx to get sin −1 x ( ) x () − −1 2 ( ) 1 u du ∫ = x sin −1 x + 1 2 ⋅ u 1 2 1 2 + C = x sin −1 x + 1 − x 2 + C 9. Use partial fractions, 9 x + 2 x + 6 ( ) x − 1 ( ) = A x + 6 + B x − 1 giving 9 x + 2 = Ax − 1 ( ) + Bx + 6 ( ) , the two equations are 9 = A + B 2 = − A + 6 B , solving gives A = 52 7 and B = 11 7 . The integral is now 52 7 x + 6 + 11 7 x − 1 ∫ dx = 52 7 ln x + 6 + 11 7 ln x − 1 + C
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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