Page 30 - Integration Problems
P. 30
Example 7:
Perform the integration
4
5
∫
3
− 1
2
The denominator is of a higher order than the
numerator, perform division.
2
_______
5
x – 1|x
3
x – x 2
5
x 2
The result is,
2
2
+
3
− 1
The integral becomes,
4 4
5 2
2
∫ = ∫ [ + ]
3
3
− 1 − 1
2 2
4 4
2
2
= ∫ + ∫
3
− 1
2 2
26
