2 (2)]
Therefore,
3y – 2 = A (y – 2) + B
Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2,
which gives A = 3 and B = 4 Therefore, the required integral is given by
I =
2
3
4
[
+
]
2
(
2)
dy
y –
y –
∫
 = 
2
3
2+ 4
(
2)
dy
dy
y –
y –
∫
∫
=
1
3log
2
4
C
2
y
–
y


−
+
+


−


=
4
3log sin
2
C
2
sin
–
φ −
+
+
φ
=
4
3log (2
sin )
+ C
2
sin
−
φ +
−
φ
 (since, 2 – sinφ is always positive)
Example 16 Find 
2
2
1
(
2) (
1)
x
x
dx
x
x
+
+
+
+
∫
Solution The integrand is a proper rational function Decompose the rational function
into partial fraction [Table 2 2(5)]