6 2 (ii)]
=
2
2
1 (
1)
2
5
2 log
1
2
5
C
2 x
x
x
x
x
x
+
+
+
+
+ +
+
+
+
Example 24 Find 
2
3
2x
x dx
−
−
∫
Solution Note that 
2
2
3
2
4
(
1)
x
x
dx
x
dx
−
−
=
−
+
∫
∫
330
MATHEMATICS
Put x + 1 = y so that dx = dy Thus
2
3
2x
x dx
−
−
∫
 =
2
4
−y dy
∫
=
2
1
1
4
4
sin
C
2
2
2
– y
y
−y
+
+
[using 7 6