Download Practice Set #3 - John Abbott College

201-203-RE - Practice Set #3: Integration by Substitution
Evaluate the following integrals.
Z
3
dx
(1)
x(ln(x))4
Z
4e3x
(2)
dx
4 + e3x
Z
8x
(3)
dx
x2 + 4
Z p
(4)
3x2 + 6x(x + 1) dx
Z
(5)
Z
(6)
Z
(10)
Z
x+4
dx
x2 + 8x
Z
p
(12)
9x2 x3 + 1 dx
(11)
(14)
3
dx
x(ln(x) + 4)4
(15)
x−3
dx
5 − 6x + x2
Z p
3
3 + ln(x)
(8)
dx
x
Z
(9)
cos(x)(2 − sin(x))4 dx
√
4
Z
e2x + x
√
dx
e2x + x2
Z
(2 + 3 ln(x))5
dx
x
(13)
8x
dx
(x2 + 2)2
Z
Z
(7)
sec2 (x)
dx
1 + 2 tan(x)
t2
Z
6x3 e2x − 5x2
dx
x3
Z
6xe3x − 18 + 5x4
dx
3x
Z
8x2 e−x + 24x + 20
dx
4x2
(16)
(17)
(18)
t+1
dt
+ 2t + 3
Z
(19)
4x1/2 e2x + 3x3 − 6
√
dx
x
x2/3 + 3x2 e3x − 5
dx
x2
Z √
3 x − 4x2 e3x
(21)
dx
3x2
Z 3x
(2 )
4
(22)
+
dx
3
3x + 1
Z
20x2 − 27x + 11
(23)
dx
4x − 3
Z 3
x + 5x
dx
(24)
x2 + 1
Z 5
3
−x
dx
(25)
3 + 3x − e
e
Z 6
4
(26)
+π
dx
3x + 5
Z
(27)
4−5x + eπ dx
Z
(20)
ANSWERS:
(1)
−1
+C
(ln(x))3
(10)
1
ln |1 + 2 tan(x)| + C
2
(20)
(2)
4
ln(4 + e3x ) + C
3
(11)
2 2
(x + 8x)3/4 + C
3
2
4
(21) − √ − e3x + C
x 9
(3) 4 ln(x2 + 4) + C
(4)
1
(3x2 + 6x)3/2 + C
9
−4
(5) 2
+C
x +2
−1
(6)
+C
(ln(x) + 4)3
(7)
1
ln |x2 − 6x + 5| + C
2
(8)
3
(3 + ln(x))4/3 + C
4
(9)
−(2 − sin(x))5
+C
5
(12) 2(x3 + 1)3/2 + C
(13) (e2x + x2 )1/2 + C
−3
5
+ e3x + + C
x
x1/3
(22)
4
23x
+ ln |3x + 1| + C
9 ln(2) 3
(14)
1
(2 + 3 ln(x))6 + C
18
(23)
(15)
1
ln(t2 + 2t + 3) + C
2
5 2
1
x − 3x + ln |4x − 3| + C
2
2
(24)
1 2
x + 2 ln(x2 + 1) + C
2
(16) 3e2x − 5 ln |x| + C
(17)
2 3x
5
e − 6 ln |x| + x4 + C
3
12
5
+C
x
√
6
(19) 2e2x + x7/2 − 12 x + C
7
(18) −2e−x + 6 ln |x| −
1
(25) −
3−x
5
− e−3x − e3 x + C
ln(3) 3
(26) 2 ln |3x + 5| + π 4 x + C
(27) −
4−5x
+ eπ x + C
5 ln(4)