Introduction to integration 327
subtraction.) Hence,
3 +
2
5
x − 6x
2
dx
= 3x +
2
5
x 1+1
1 + 1
− (6)
x 2+1
2 + 1
+ c
= 3x +
2
5
x 2
2
− (6)
x 3
3
+ c = 3x +
1
5
x
2
− 2x
3
+ c
Note that when an integral contains more than one term
there is no need to have an arbitrary constant for each;
just a single constant c at the end is sufficient.
Problem 6. Determine
2x 3 − 3x
4x
dx
Rearranging into standard integral form gives
2x 3 − 3x
4x
dx =
2x 3
4x
−
3x
4x
dx
=
1
2
x
2
−
3
4
dx =
1
2
x 2+1
2 + 1
−
3
4
x + c
=
1
2
x 3
3
−
3
4
x + c =
1
6
x
3
−
3
4
x + c
Problem 7. Determine
1 − t
2 dt
Rearranging
(1 − t )
2 dt gives
(1 − 2t + t
2
)dt = t −
2t 1+1
1 + 1
+
t 2+1
2 + 1
+ c
= t −
2t 2
2
+
t 3
3
+ c
= t − t
2
+
1
3
t
3
+ c
This example shows that functions often have to be rearranged into the standard form of
ax n dx before it is
possible to integrate them.
Problem 8. Determine
5
x 2 dx
5
x 2 dx =
5x
−2 dx
Using the standard integral,
ax
n dx, when a = 5 and
n = −2, gives
5x
−2 dx =
5x −2+1
−2 + 1
+ c =
5x −1
−1
+ c
= −5x
−1
+ c = −
5
x
+ c
Problem 9. Determine
3
√
xdx
For fractional powers it is necessary to appreciate
n
√
a m = a
m
n
3
√
x dx =
3x
1
2 dx =
3x
1
2 +1
1
2
+ 1
+ c =
3x
3
2
3
2
+ c
= 2x
3
2 + c = 2
x 3 + c
Problem 10. Determine
−5
9
4
√
t 3
dt
−5
9
4
√
t 3
dt =
−5
9t
3
4
dt =
−
5
9
t
−
3
4 dt
=
−
5
9
t
−
3
4 +1
−
3
4
+ 1
+ c
=
−
5
9
t
1
4
1
4
+ c =
−
5
9
4
1
t
1
4 + c
= −
20
9
4
√
t + c
Problem 11. Determine
4 cos 3x dx
From 2 of Table 35.1,
4 cos 3x dx = (4)
1
3
sin 3x + c
=
4
3
sin 3x + c
Problem 12. Determine
5 sin2θdθ