Therefore,
I1=  (
)
2
4
2
5
4
–
x dx
dt
t
x
x
−
=
−
−
∫
∫
 = 
1
2
C
t +
=
2
1
2 5 4
– x – x +C (2)
Now consider
I2 =
2
2
5
4
9
(
2)
dx
dx
x
x
– x
=
−
−
+
∫
∫
Put x + 2 = t, so that dx = dt Therefore,
I2 =
1
2
2
2
sin
3+ C
3
–
dt
t
t
=
−
∫
[by 7 4 (5)]
=
1
2
2
sin
C
3
– x +
+