
14-16 Calculus – Differentiation and Integration
Case 3: Assume that b ac
2
4 0− > and a is negative.
dx
ax bx c
dx
a
b ac
a
x
b
a
2
2
2
2
4
2 2
+ +
∫
=
−
− +
∫
,
wwhich is in the form
dx
a x
a
ax b
b ac
2 2
1
2
1 2
4
−
∫
= −
+
−
−
sin
,
where is positivea .
Example 14.23 Evaluate
dx
x x2 3 4
2
+ +
∫
.
Solution: Here, a = 2, b = 3, c = 4
Also b
2
-4ac = -23 < 0 and a is positive
dx
x x
dx
x
x
2 3 4
3
4
23
4
1
2
2
2
2
1
+ +
∫
=
+
+
∫
=
−
sinh
++
=
+
−
3
4
23 4
1
2
4 3
2 23
1
/
sinh
x
.
Example 14.24 Evaluate
dx
x x8 4 4
2
− −
∫
.
Solution: Here, a = -4, b = -4, c = 8.
And
b ac
2
4 144 0− = > and a is negative.
Thus,
dx
ax bx c
a
ax b