4 (10) Let us express
x + 3 = 
2
A
(5
4
) + B
d
– x – x
dx
= A (– 4 – 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
– 2A = 1 and – 4 A + B = 3, i 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