
Integration of Irrational Functions 14-21
Reduction Formula for Evaluation of
dx
Ax B ax bx c( )+ + +
∫
2
Let Ax B
t
dx
dt+ = = −
1 1
2
, .then
Also x
tB
tA
=
−1
.
Now,
dx
Ax B ax bx c
t
At
dt
a
tB
tA
b
( )+ + +
∫
=
−
−
+
−
2
2
2
1
1 1
ttB
tA
c
dt
a tB btA tB cA t
dt
aB cA
+
∫
= −
− + − +
∫
= −
+
( ) ( )
(
1 1
2 2 2
2
22 2
2− + − +
∫
bAB t bA aB t a) ( )
,
where t
Ax B
=
+
1
,
which we can easily evaluate using the standard formula.
Example 14.33 Evaluate
1
1 2 4 1
2
( )
.
x x x
dx
− − +
∫
Solution: Here, a = 2, b = -4, c = 1, A = 1 and B = -1.
Also b
2
- 4ac = 8 > 0 and a is positive.
We have
dx
Ax B ax bx c
dt
aB cA bAB t bA aB t a( ) ( ) ( )
,
+ + +
∫
= −
+ − + − +
∫
2 2 2 2
2
where t
Ax B
=
+
1