Tutorial 4: Integration
ECO4112F 2011
(x + 1)20 dx
R √
2. x x2 + 5dx
2
R
3. 4x2 x4 + 1 dx
R
2
4. 2xex dx
3
R 2
5.
x + 1 ex +3x dx
1.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
R
19.
R3
3x2 − x + 6 dx
−1
20.
R1
e3t dt
0
21.
R1
2x2 x3 − 1
3
dx
0
R 7
dx
x
Z
2x3 + 3x
dx
x4 + 3x2 + 7
R√
2x − 1dx
R −x/4
e
dx
2
R 2
x + 1 dx
Z √x
e
√ dx
x
Z 1
1
+
dx
x − 1 (x − 1)2
R
sin (5x − 3) dx
Z
7
dx
3 − 2x
R 3−x
2 dx
Z
9x2 + 5
dx
3x
Z ln x
3
dx
x
Z
x+3
dx
x+6
22.
R1 √
3
x5 dx
−1
23.
R√
Z
24.
25.
R2
103x dx
ln x
√ dx
x
x ln xdx
1
2
26.
R
xex dx
27.
R
x2 e2x+1 dx
28.
R
xe−x dx
y 3 ln ydy
R √
30. x x + 1dx
29.
R
31.
R2
32.
R
(2x − 1) ln (x − 1) dx
33.
R
ex sin xdx
34.
Re √
xe2x dx
1
1
1
x ln x2 dx
Tutorial 4: Integration
SELECTED SOLUTIONS
ECO4112F 2011
1.
R
(x + 1)20 dx
Easy
R p
2. x x2 + 5dx
3=2
x2 + 5
=
+C
3
R
2
3. 4x2 x4 + 1 dx
x3
11
3
x11
=4
4.
5.
6.
7.
8.
9.
10.
R
+
2x7
7
+
+C
14.
2
=
x5
5
+
2x3
3
p
x
1
x
...
R
+C
1
+
sin (5x
Z
1
(x
1)2
dx
3) dx
cos (5x
5
3)
7
dx
3 2x
Simple, but be careful of signs
R
15. 23 x dx
1 3 x
=
2
+C
ln 2
Z
9x2 + 5
16.
dx
3x
Hint: divide through
Z ln x
3
17.
dx
x
2
2
p
e x
p dx
x
=
2xex dx
x2 + 1
12.
= 2e
Z
13.
= ex + C
R 2
3
x + 1 ex +3x dx
1 3
= ex +3x + C
3
R 7
dx
x
Easy
Z
2x3 + 3x
dx
x4 + 3x2 + 7
Easy
Rp
2x 1dx
1
= (2x 1)3=2 + C
3
R x=4
e
dx
...
R
11.
Z
=
18.
dx
19.
Z
3ln x
+C
ln 3
x+3
dx
x+6
Hint: rewrite the top and break up into
two fractions ...
R3
3x2
1
+x+C
= 48
1
x + 6 dx
20.
R1
27.
e3t dt
0
1 3
=
e
3
21.
R1
=
1
28.
2x2 x3
1
3
22.
1
6
R1 p
3
x5 dx
103x dx
p
2 103x
+C
=
3 ln 10
Z
ln x
p dx
24.
x
p
= 2 x (ln x 2) + C
25.
31.
R
+C
x dx
R2
xe2x dx
1
= e2 3e2 1
4
R
32. (2x 1) ln (x
= x (x
x ln xdx
= 2 ln 2
1
2
1
1
26.
xe
x+
1
y4
ln y
+C
4
4
R p
30. x x + 1dx
2
(x + 1)3=2 (3x 2) + C
=
15
=0
Rp
R2
e2x+1 2
x
2
=
1
23.
R
x2 e2x+1 dx
Easy, use by parts
R
29. y 3 ln ydy
dx
0
=
R
33.
3
4
R
1) ln (x
1) dx
1)
x2
+C
2
ex sin xdx
Don’t use by parts more than twice
2
xex dx
1 2
= ex + C
2
34.
Re p
x ln x2 dx
1
Use by parts.
2
ECO4112F, Tutorial 4
12
ˆ "
#
1
1
+
dx
x − 1 (1 − x)2
ˆ
=
ˆ
ln x
√ dx
x
ˆ
1
−1
ˆ
=
=
=
=
=
1
(ln x) x− 2 dx
ˆ
1 1
1
2
2x ln x + A − 2x 2 dx
x
ˆ
1
1
2x 2 ln x + A − 2x− 2 dx
1
1
2x 2 ln x + A − 4x 2 − B
√
2 x (ln x − 2) + C
25
ˆ
2
x ln xdx
1
34
ˆ
e
√
2 ˆ 2
1 21
1 2
x ln x −
x dx
=
2
x
1 2
1
2
1 2 1
=
2 ln 2 − ln 1 − x 2
4 1
3
= 2 ln 2 −
4
ˆ
2
x ln x dx
=
e
1
2x 2 ln xdx
1
e ˆ e
4 3
4 31
2
x ln x −
x 2 dx
3
3
x
1
1
e
4 3
8 3
e 2 − x 2 3
9
1
4 3
8 3
8
e2 − e2 +
3
9
9
4 3
8
e2 +
9
9
1
=
=
=
=
1
2 dx
(x − 1)
ln (x − 1) + − (x − 1)
=
24
1
dx +
x−1
+C