4 (7) We write the denominator of the integrand,
32
13
10
x
x –
+
 =
2
13
10
3
3
3
x
x
–


+




=
2
2
13
17
3
6
6
x
–






+














(completing the square)
Thus
3
13
10
dx
x
x
2 +
−
∫
 =
2
2
31
13
17
6
6
dx
x



+
−








∫
Put 
13
6
x
t
+
= Then dx = dt Therefore,
3
13
10
dx
x
x
2 +
−
∫
 =
2
2
31
17
6
dt
t


− 



∫
=
1
17
1
6
log
C
17
17
3
2
6
6
t –
t
+
× ×
+
[by 7