e , A = 
1
2
–
 and B = 1
INTEGRALS         315
Therefore,
2
3
5
4
x
dx
x
x
+
−
−
∫
 =
(
)
2
2
4
2
21
5
4
5
4
–
– x dx
dx
–
x
x
x
x
+
−
−
−
−
∫
∫
=
–21
 I1 + I2 (1)
In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt 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