QUIZ 7 - SOLUTIONS
1. Find
!
x ln(x) dx
2
Answer: Let u = ln(x) and dv = xdx, and then du = dx
and v = x2 .
x
!
! 2
x2
x 1
x ln(x) dx = ln(x) −
· dx
2
2 x
!
x2 ln(x)
x
=
−
dx
2
2
x2 ln(x) x2
=
−
+C
2
4
2. Find
!
3x sin(
πx
) dx
7
Answer: Let u = 3x and dv = sin( πx
) dx, and then du = 3 dx and v = − π7 cos( πx
).
7
7
"
# !
!
πx
7
πx
7
πx
3x sin( ) dx = 3x − cos( ) − − cos( ) 3 dx
7
π
7
π
7
!
21x
πx
21
πx
=−
cos( ) +
cos( ) dx
π
7
π
7
21x
πx
21 7
πx
=−
cos( ) +
· sin( ) + C
π
7
π π
7
21x
πx
147
πx
=−
cos( ) + 2 sin( ) + C
π
7
π
7
QUIZ 8 - SOLUTIONS
1. Find
!
9
(4x + 1) ln(2x + 1) dx
1
2
Answer: Let u = ln(2x + 1) and dv = (4x + 1)dx, and then du = 2x+1
dx and v = 2x2 + x.
! 9
! 9
"9
2
2
"
(4x + 1) ln(2x + 1) dx = (2x + x) ln(2x + 1) 1 −
(2x2 + x)
dx
2x + 1
1
1
! 9
"9
2
2
dx
= (2x + x) ln(2x + 1)"1 −
x(2x + 1)
2x + 1
1
! 9
"9
2
"
= (2x + x) ln(2x + 1) 1 −
2x dx
1
"9
= (2x2 + x) ln(2x + 1) − x2 "1
#
$ #
$
= (2 · 92 + 9) ln(2 · 9 + 1) − 92 − (2 · 12 + 1) ln(2 · 1 + 1) − 12
= 171 ln(19) − 3 ln(3) − 80
2. Find
!
4 √
e
x
dx
1
Answer: First a substition. Let w =
!
√
1
x and dw = 2√1 x dx, and then 2w dw = dx.
! 2
4 √
x
e dx =
2wew dw
1
Then we do integration by parts. Let u = 2w and dv = ew dw, and then du = 2 dw
and v = ew .
! 4 √
! 2
x
2wew dw
e dx =
1
1
! 2
w 2
= 2we |1 −
2ew dw
w
= 2we −
1
w 2
2e |1
2
= (2 · 2e2 − 2e ) − (2 · 1e1 − 2e1 )
= 2e2