, to
indicate that the statement is true only for certain values of x Example 12 Find 
2
2
1
5
6
x
dx
x
−+x
+
∫
Solution Here the integrand 
2
2
1
5
6
x
x – x
+
+
 is not proper rational function, so we divide
x2 + 1 by x2 – 5x + 6 and find that
INTEGRALS         319
2
2
1
5
6
x
x – x
+
+
 =
2
5
5
5
5
1
1
(
2) (
3)
5
6
x –
x –
x –
x –
+x – x
= +
+
Let
5
5
(
2) (
3)
x –
x –
x –
 =
A
B
2
3
x –
x –
+
So that
5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5
and 3A + 2B = 5 Solving these equations, we get A = – 5  and B = 10
Thus,
2
2
1
5
6
x
x – x
+
+
 =
5
10
1
2
3
x –
x –
−
+
Therefore,
2
2
1
5
6
x
dx
x – x
+
+
∫
 =
1
5
10
2
3
dx
dx
dx
x –
x –
−
+
∫
∫
∫
= x – 5 log |x – 2| + 10 log |x – 3| + C Example 13 Find 
32
2
(
1) (
3)
x
dx
x
−x
+
+
∫
Solution The integrand is of the type as given in Table 7