Chapter – 14
Antiderivatives

Exercise 14.1

Standard I

∫
dx x2 + a2
  =  
1a
tan-1 (
xa
) + C
∫
dx x2 - a2
  =  
12a
log
x - ax + a
+ C
∫
dx a2 - x2
  =  
12a
log (
a + xa - x
) + C
(Here, x < a)

Exercise 14.1

(i) ∫
dx x2 + 49
=
∫
dx (x)2 + (7)2
=
17
tan-1 (
x7
) + C
∵ ∫
dx x2+a2
=
1a
tan-1(
xa
) + C

Class Work:

CW1) ∫
dx x2 - 16


CW2) ∫
dx 25 - x2




CW1) ∫
dx x2 - 16
=
∫
dx (x)2 - (4)2
=
12(4)
log
x - 4x + 4
+ C
∵ ∫
dx x2-a2
=
12a
log
x-ax+a
=
18
log (
x - 4x + 4
) + C
CW2) ∫
dx 25 - x2
=
∫
dx (5)2 - (x)2
=
12(5)
log (
5 + x5 - x
) + C
∵ ∫
dx a2-x2
=
12a
log
a+xa-x
=
110
log (
5 + x5 - x
) + C

Question (ii)

(ii) ∫
dx 4x2 + 25
=
∫
dx (2x)2 + (5)2
Let:
y = 2x
Differentiating on both sides w.r.t x
dydx
= 2   ⇒   dx =
12
dy
Substituting the values of x and dx:
=
∫
12
dy y2 + (5)2
=
12
∫
dy y2 + (5)2
=
12
· [
15
tan-1(
y5
) ] + C
∵ ∫
dx x2+a2
=
1a
tan-1(
xa
)
=
110
tan-1(
2x5
) + C
Alternative: Direct Standard Form
∫
dx (2x)2 + (5)2
 = 
110
tan-1(
2x5
) + C
Note: Direct method is faster, but substitution is clearer and more systematic, helping to understand and solve complex problems later.

Simple Rule

If the numerator is exactly the derivative of the denominator:
∫
derivative of denominator denominator
dx  =  log( denominator ) + C

Examples

1. ∫
1x
dx  =  log(x) + C
2. ∫
2xx2 + 1
dx  =  log(x2 + 1) + C
3. ∫
3x2x3 + 5
dx  =  log(x3 + 5) + C
4. ∫
1x + 4
dx  =  log(x + 4) + C
5. ∫
exex + 2
dx  =  log(ex + 2) + C
6. ∫
cos xsin x
dx  =  log(sin x) + C
7. ∫
xx2 + 9
dx  = 
12
· ∫
2xx2 + 9
dx  = 
12
log(x2+9)+C
8. ∫
4x3x4 - 3
dx  =  log(x4 - 3) + C
9. ∫
1x log x
dx  =  ∫
1x
 log x
dx  =  log(log x) + C
10. ∫
5x4x5 - 2
dx  =  log(x5 - 2) + C
TIP : Before applying Standard Integral forms, always first check if you can apply this rule or not !!

Question (iv)

(iv) ∫
2x + 3 4x2 + 1
dx
Differentiation of denominator (4x2+1) is 8x.
=
∫
14
(8x) + 3 4x2 + 1
dx
=
14
∫
8x 4x2 + 1
dx  +  3 ∫
dx 4x2 + 1
=
14
log(4x2 + 1)  +  3 I1    ...... (1)
∵ ∫
f'(x) f(x)
= log f(x)
Solving I1:
I1 = ∫
dx 4x2 + 1
 =  ∫
dx (2x)2 + (1)2
Let:
y = 2x
Diff w.r.t x:
dx =
12
dy
I1 = ∫
12
dy y2 + 12
 = 
12
[
11
tan-1(
y1
) ]
I1 =
12
tan-1(2x)
Substituting I1 back into equation (1):
∫ ... dx  = 
14
log(4x2 + 1)  +  3 [
12
tan-1(2x) ] + C
=
14
log(4x2 + 1) +
32
tan-1(2x) + C

Question 5)

5) ∫
6x + 1 x2 + 9
dx
Differentiation of denominator (x2+9) is 2x.
Break 6x into 3 × 2x to match the derivative.
=
∫
3(2x) + 1 x2 + 9
dx
=
3 ∫
2x x2 + 9
dx  +  ∫
dx x2 + 9
=
3 log(x2 + 9)  +  I1    ...... (1)
Solving I1:
I1 = ∫
dx x2 + 9
=
∫
dx x2 + 32
Using:
1a
tan-1(
xa
)
I1 =
13
tan-1(
x3
)
Substituting I1 back into equation (1):
=
3 log(x2 + 9)  + 
13
tan-1(
x3
) + C

Question (iii)

(iii) ∫
x x4 + 3
dx
=
∫
x dx (x2)2 + 3
Let:
y = x2
Differentiating on both sides w.r.t x
dydx
= 2x   ⇒   x dx =
12
dy
Substituting the values:
=
∫
12
dy y2 + 3
=
12
∫
dy (y)2 + (√3)2
=
12
· [
1√3
tan-1(
y√3
) ] + C
∵ ∫
dx x2+a2
=
1a
tan-1(
xa
)
=
12√3
tan-1(
x2√3
) + C

Question (vi)

(vi) ∫
dx ex + e-x
=
∫
dx ex +
1ex
=
∫
ex dx (ex)2 + 1
Let:
y = ex
Differentiating on both sides w.r.t x
dydx
= ex   ⇒   ex dx = dy
Substituting the values:
=
∫
dy (y)2 + (1)2
=
11
tan-1(
y1
) + C
∵ ∫
dx x2+a2
=
1a
tan-1(
xa
)
=
tan-1(ex) + C

Topic: Completing the Square

Reference Formulas:
(a + b)2 = a2 + 2ab + b2 (a - b)2 = a2 - 2ab + b2
1) x2 + 6x + 8
=
(x)2 + 2 · x · 3 + 8
=
(x)2 + 2 · x · 3 + (3)2 - (3)2 + 8
=
(x + 3)2 - 9 + 8
=
(x + 3)2 - 1
2) x2 + 4x - 12
=
(x)2 + 2 · x · 2 - 12
=
(x)2 + 2 · x · 2 + (2)2 - (2)2 - 12
=
(x + 2)2 - 4 - 12
=
(x + 2)2 - 16
3) sin2 x + 4 sin x + 5
=
(sin x)2 + 2 · sin x · 2 + 5
=
(sin x)2 + 2 · sin x · 2 + (2)2 - (2)2 + 5
=
(sin x + 2)2 - 4 + 5
=
(sin x + 2)2 + 1
4) 1 + x - x2
=
- [ x2 - x - 1 ]
=
- [ (x)2 - 2 · x ·
12
 - 1 ]
=
- [ (x)2 - 2 · x ·
12
 + (
12
)2 - (
12
)2 - 1 ]
=
- [ (x -
12
)2 -
14
- 1 ]
=
- [ (x -
12
)2 -
54
]
=
54
- (x -
12
)2

Question (ix)

(ix) ∫
dx x2 + 6x + 8
Complete the square in the denominator.
=
∫
dx (x)2 + 2·x·3 + (3)2 - (3)2 + 8
=
∫
dx (x + 3)2 - 9 + 8
=
∫
dx (x + 3)2 - 1
Let:
y = x + 3   ⇒   dy = dx
Substituting values:
=
∫
dy (y)2 - (1)2
=
12(1)
log
y - 1 y + 1
+ C
∵ ∫
dx x2-a2
=
12a
log
x-a x+a
=
12
log
(x + 3) - 1 (x + 3) + 1
+ C
=
12
log (
x + 2 x + 4
) + C

Question (xi)

(xi) ∫
dx 1 + x - x2
Step 1: Factor out the negative sign from the denominator to make x2 positive.
=
∫
dx - [ x2 - x - 1 ]
=
∫
dx - [ (x)2 - 2·x·
12
 + (
12
)2 - (
12
)2 - 1 ]
=
∫
dx - [ (x -
12
)2 -
54
]
=
∫
dx (
√52
)2 - (x -
12
)2
Let:
y = x -
12
  ⇒   dy = dx
Substituting values:
=
∫
dy (
√52
)2 - (y)2
∵ ∫
dx a2-x2
=
1 2 · (
√52
)
log [
√52
+ y
√52
- y
] + C
=
1√5
log [
√52
+ (x -
12
)
√52
- (x -
12
)
] + C
=
1√5
log (
√5 + 2x - 1 √5 - 2x + 1
) + C

Question (xiii)

(xiii) ∫
x x2 + 4x - 12
dx
Derivative of denominator is 2x + 4.
=
∫
12
(2x + 4) - 2 x2 + 4x - 12
dx
=
12
∫
2x + 4 x2 + 4x - 12
dx  −  2 ∫
dx x2 + 4x - 12
=
12
log(x2 + 4x - 12)  −  2 ∫
dx (x + 2)2 - 16
=
12
log(x2 + 4x - 12)  −  2 ∫
dx (x + 2)2 - (4)2
=
12
log(x2 + 4x - 12)  −  2 · [
1 2(4)
log
(x + 2) - 4 (x + 2) + 4
] + C
∵
12a
log
x-ax+a
=
12
log(x2 + 4x - 12)  − 
28
log
x - 2 x + 6
+ C
=
12
log(x2 + 4x - 12)  − 
14
log (
x - 2 x + 6
) + C

Question (xvi)

(xvi) ∫
2x + 2 3 + 2x - x2
dx
=
∫
2x + 2 - (x2 - 2x - 3)
dx  =  - ∫
2x + 2 x2 - 2x - 3
dx
Derivative of denominator (x2 - 2x - 3) is 2x - 2.
=
- ∫
(2x - 2) + 4 x2 - 2x - 3
dx
=
- [ ∫
2x - 2 x2 - 2x - 3
dx  +  4 ∫
dx x2 - 2x - 3
]
=
- ∫
2x - 2 x2 - 2x - 3
dx  −  4 ∫
dx x2 - 2x - 3
=
- log(x2 - 2x - 3)  −  4 I1
Solving I1 (Completing the Square):
x2 - 2x - 3  =  x2 - 2(x)(1) + 12 - 12 - 3  =  (x - 1)2 - 4  =  (x - 1)2 - 22
I1 = ∫
dx (x - 1)2 - 22
Using:
12a
log
x-ax+a
=
12(2)
log
(x - 1) - 2 (x - 1) + 2
=
14
log
x - 3 x + 1
Putting it back together:
=
- log(x2 - 2x - 3)  −  4 [
14
log
x - 3 x + 1
] + C
=
- log(x2 - 2x + 3) - log
x - 3 x + 1
+ C