Coroneos' 100 Integrals - · PDF fileChapter 1 The 100 Integrals The ideal way to use this book is to: • Do a section of ten in one go. (do NOT look at the solutions) • Mark your

CORONEOS’ 100 INTEGRALS CHALLENGE

PrefaceThe art of Integration is an extremely important skill to master in the HSC. With heavy use in Volumes, resisted motion in Mechanics, and having its own standalone topic, if you’re not good at Integration, you’re going to have a bad time in 4U maths.

Coroneos’ 100 Integral challenge is known for having a rather comprehensive coverage all the type of integrals you may encounter in the HSC. If you are able to complete all 100 integrals with ease, then no integral in the HSC should cause you great difficulty.

Why Does This Book Exist?

There’s no point in learning the trick/manipulation/technique/answer to a particular question if you don’t understand how to use it in other questions. So alongside the worked solutions, we’ve also tried to teach you HOW to approach the question, and questions similar to it.

Chapter 1

The 100 Integrals

The ideal way to use this book is to:

• Do a section of ten in one go. (do NOT look at the solutions)

• Mark your answers using the quick solutions.

• Carefully read the worked solutions provided. Ask yourself:

- Was my method the most efficient one?

- Why have the solutions done what they have done?

- What would I change if I were doing this integral again?

- How should I approach this type of integral in the future?

• Repeat the process for the next set of ten integrals.

1. ∫x

x2 + 4d x

2. ∫x

x2 + 4d x

3. ∫5x + 2x2 − 4

d x

4. ∫ sin x cos3 x d x

5. ∫ sin x sec3 x d x

6. ∫ cos2 x2

d x

7. ∫ x sin x d x

8. ∫ x sec2 2x d x

9. ∫ tan−1 2x d x

10. ∫x3

x2 + 1d x

Section 1

3

Questions 1 - 10

11. ∫x

(x + 2)(x + 4)d x

12. ∫(x − 1)(x + 1)(x − 2)(x − 3)

d x

13. ∫2x − 1

x2 + 2x + 3d x

14. ∫x3

2x − 1d x

15. ∫1 + x

1 − x − x2d x

16. ∫d x

x2(1 − x2)12

17. ∫d x

x a2 + x2

18. ∫d x

x a2 − x2

19. ∫d x

x x2 − a2

20. ∫x

x + 1d x

Section 2

4

Questions 11 - 20

21. ∫cos−1 x

1 − x2d x

22. ∫x + 1x − 1

d x

23. ∫d x

x(ln x)3

24. ∫ sec4 3x d x

25. ∫d x

x2(1 − x)

26. ∫d x

x2(1 + x2)

27. ∫d x

(1 + x2)2

28. ∫ tan3 x d x

29. ∫d x

5 + 3 cos x

30. ∫d x

3 + 5 cos x

Section 3

5

Questions 21 - 30

31. ∫sin x

5 + 3 cos xd x

32. ∫d x

1 + cos2 x

33. ∫d x

cos2 x2 − sin2 x

2

34. ∫ x2 sin x d x

35. ∫x2

(x − 1)(x − 2)(x − 3)d x

36. ∫ex

ex − 1d x

37. ∫d x

3 sin2 x + 5 cos2 x

38. ∫ x3e5x4−7 d x

39. ∫ x5 ln x d x

40. ∫3x + 2

x(x + 1)3d x

Section 4

6

Questions 31 - 40

41. ∫ ln(x3) d x

42. ∫d x

ex + e−x

43. ∫ (5x3 + 7x − 1)32 ⋅ (15x2 + 7) d x

44. ∫d x

(x2 + 1)(x2 + 4)

45. ∫ (x2 + x + 1)−1 d x

46. ∫ ex sin 2x d x

47. ∫ (x2 + x − 1)−1 d x

48. ∫ (x2 − x)− 12 d x

49. ∫1 − 2x3 + x

d x

50. ∫ x3(4 + x2)− 12 d x

Section 5

7

Questions 41 - 50

51. ∫sin 2x

3 cos2 x + 4 sin2 xd x

52. ∫x2

1 − x4d x

53. ∫d x

sin x cos x

54. ∫ ln x − 1 d x

55. ∫d x

ex − 1

56. ∫sec2 x

tan2 x − 3 tan x + 2d x

57. ∫x + 1

(x2 − 3x + 2)12

d x

58. ∫ sin 2x cos x d x

59. ∫x

1 + x3d x

60. ∫ x tan−1 x d x

Section 6

8

Questions 51 - 60

61. ∫ (1 + 3x + 2x2)−1 d x

62. ∫ (9 − x2)12 d x

63. ∫ (9 + x2)12 d x

64. ∫ x(9 + x2)12 d x

65. ∫ sec2 x tan3 x d x

66. ∫ x2e−x d x

67. ∫ xex2 d x

68. ∫ sin x tan x d x

69. ∫ sin4 x cos3 x d x

70. ∫x3 + 1x3 − x

d x

Section 7

9

Questions 61 - 70

71. ∫ ln(x + x2 − 1) d x

72. ∫d x

(x + 1)12 + (x + 1)

73. ∫4

0

x

x + 4d x

74. ∫2

1

d xx(1 + x2)

75. ∫2

1

ln xx

d x

76. ∫1

0cos−1 x d x

77. ∫2

1

x + 1

−2 + 3x − x2d x

78. ∫π2

0

d xcos2 x + 2 sin2 x

79. ∫1

0x 1 − x2 d x

80. ∫4

2x ln x d x

Section 8

10

Questions 71 - 80

81. ∫2

1

d xx2 + 5x + 4

82. ∫π2

0 (1 +12

sin x)−1

d x

83. ∫1

0x2e−x d x

84. ∫1

0

7 + x1 + x + x2 + x3

d x

85. ∫1

0

e−2x

e−x + 1d x

86. ∫a2

0

ya − y

dy

87. ∫a

0

(a − x)2

a2 + x2d x

88. ∫1

0

x + 3(x + 2)(x + 1)2

d x

89. ∫1

0

x2

x6 + 1d x

90. ∫π

0cos2 m x d x, where m is an integer.

Section 9

11

Questions 81 - 90

91. ∫π2

π4

x sin 2x d x

92. ∫a2

0x2 a2 − x2 d x

93. ∫π4

0sec2 x tan x d x

94. ∫1

0(x + 2) x2 + 4x + 5 d x

95. ∫2

1x(ln x)2 d x

96. ∫4

3

x2 + 4x2 − 1

d x

97. ∫4

1

x2 + 4x(x + 2)

d x

98. ∫π2

0

cos x5 − 3 sin x

d x

99. ∫1

0

1

(4 − x2)32

d x

100.∫π2

02 sin θ cos θ(3 sin θ − 4 sin3 θ) dθ

Section 10

12

Questions 91 - 100

Chapter 2

Quick Solutions

1. 12

ln |x2 + 4 | + C

2. x2 + 4 + C

3. 2 ln |x + 2 | + 3 ln |x − 2 | + C

4. −cos4 x

4+ C

5. 12

sec2 x + C

6. x2

+sin x

2+ C

7. −x cos x + sin x + C

8. x2

tan 2x +14

ln |cos 2x | + C

9. x tan−1 2x −14

ln |1 + 4x2 | + C

10. x2

2−

12

ln |x2 + 1 | + C

14

Questions 1 - 100

WAIT!Are you sure you should be here?

We strongly advise you to attempt questions in blocks of ten. And dont check your answers until you finish the block!

Don’t use the solutions to give you hints; you’re only cheating yourself. Make sure you give each question your best attempt

before coming to this section (or the next) to check your answers!

11. 2 ln |x + 4 | − ln |x + 2 | + C

12. x + 8 ln |x − 3 | − 3 ln |x − 2 | + C

13. ln |x2 + 2x + 3 | −3

2tan−1 ( x + 1

2 ) + C

14. x3

6+

x2

8+

x8

+116

ln |2x − 1 | + C

15. − 1 − x − x2 +12

sin−1 ( 2x + 1

5 ) + C

16. −1 − x2

x+ C

17. −1a

lnax

+a2 + x2

x+ C

18. −1a

lnax

+a2 − x2

x+ C

19. 1a

sec−1 xa

+ C

20. 23

x32 − x + 2x

12 − 2 ln ( x + 1) + C

15

Questions 11 - 20

21. −(cos−1 x)2

2+ C

22. x2 − 1 + ln x + x2 − 1 + C

23. −1

2(ln x)2+ C

24. tan 3x3

+tan3 3x

9+ C

25. lnx

1 − x−

1x

+ C

26. −1x

− tan−1 x + C

27. tan−1 x2

+x

2(x2 + 1)+ C

28. tan2 x2

+ ln |cos x | + C

29. 12

tan−1 (tan x

2

2 ) + C

30. 14

ln2 + tan x

2

2 − tan x2

+ C

16

Questions 21 - 30

31. −13

ln |5 + 3 cos x | + C

32. =1

2tan−1 ( tan x

2 ) + C

33. ln |sec x + tan x | + C

34. −x2 cos x + 2x sin x + 2 cos x + C

35. 12

ln |x − 1 | − 4 ln |x − 2 | +92

ln |x − 3 | + C

36. ln |ex − 1 | + C

37.15

15tan−1 ( 3

5tan x) + C

38. 120

e5x4−7 + C

39. x6

6ln x −

x6

36+ C

40. 2 ln |x | − 2 ln |x + 1 | +2

x + 1−

12(x + 1)2

+ C

17

Questions 31 - 40

41. 3x ln x − 3x + C

42. tan−1(ex) + c

43. 25

(5x3 + 7x − 1)52 + C

44. 13

tan−1 x −16

tan−1 ( x2 ) + C

45. 2

3tan−1 ( 2x + 1

3 ) + C

46. ex

5 (sin 2x − 2 cos 2x) + C

47. 1

5ln

2x + 1 − 5

2x + 1 + 5+ C

48. ln 2x − 1 + 2 x2 − x + C

49. 7 ln |x + 3 | − 2x + C

50. 13

(x2 − 8) x2 + 4 + C

18

Questions 41 - 50

51. ln(3 + sin2 x) + C

52. 14

ln |1 + x | −14

ln |1 − x | −12

tan−1 x + C

53. −2 ln |cosec x + cot x | + C OR = ln | tan x | + C

54. 12

(x − 1)ln(x − 1) −12

x + C

55. ln |1 − e−x | + C OR ln |ex − 1 | − x + C

56. ln | tan x − 2 | − ln | tan x − 1 | + C

57. x2 − 3x + 2 +52

ln 2x − 3 + 2 x2 − 3x + 2 + C

58. −23

cos3 x + C OR − 16

cos 3x −12

cos x + C

59. −13

ln |1 + x | +16

ln |1 − x + x2 | +1

3tan−1 ( 2x − 1

3 ) + C

60. x2

2tan−1 x −

12

x +12

tan−1 x + C

19

Questions 51 - 60

61. ln |2x + 1 | − ln |x + 1 | + C

62. 12 [9 sin−1 ( x

3 ) + x 9 − x2] + C

63. 12

x x2 + 9 +92

ln x + x2 + 9 + C2

64. 13

(x2 + 9)32 + C

65. tan4 x4

+ C

66. −e−x(x2 + 2x + 2) + C

67. 12

ex2 + C

68. ln |sec x + tan x | − sin x + C

69. 15

sin5 x −17

sin7 x + C

70. x − ln |x | + ln |x − 1 | + C

20

Questions 61 - 70

71. x ln (x + x2 − 1) − x2 − 1 + C

72. 2 ln x + 1 + 1 + C

73. 163 (2 − 2)

74. 12

ln85

75. 12

(ln 2)2

76. 1

77. 5π2

78.π 2

4

79. 13

80. 14 ln 2 − 3

21

Questions 71 - 80

81. 13

ln ( 54 )

82. 2π

3 3

83. 2 − 5e−1

84. 32

ln 2 + π

85. ln ( 1 + e2e ) −

1e

+ 1

86. a2

(ln 4 − 1)

87. a(1 − ln 2)

88. ln34

+ 1

89. π12

90. π2

22

Questions 81 - 90

91. π4

−14

92. a4

192 (4π − 3 3)

93. 12

94.5 5

3 (2 2 − 1)

95. 2(ln 2)2 − 2 ln 2 +34

96. 1 +52

ln65

97. 3

98. 13

ln52

99.3

12

100. 25

23

Questions 91 - 100

Chapter 3

Worked Solutions

When reading the worked solutions provided, ask yourself:

- Was my method the most efficient one?

- Why have the solutions done what they have done?

- What would I change if I were doing this integral again?

- How should I approach this type of integral in the future?

This is the most important part when studying for Integration. Doing the integrals themselves does absolutely nothing. Your studying only truly begins once you carefully examine the worked solutions and reflect upon what you would do differently next time.

1. ∫x

x2 + 4d x

Method 1: Reverse Chain Rule

Observing that dd x

(x2 + 4) = 2x:

∫x

x2 + 4d x =

12 ∫

2xx2 + 4

d x

=12

ln |x2 + 4 | + C, applying the reverse chain rule.

Method 2: Algebraic Substitution

Observing that dd x

(x2 + 4) = 2x, u = x2 + 4 is a suitable choice of substitution:

∫x

x2 + 4d x = ∫

1u ( 1

2du)

=12 ∫

duu

=12

ln |u | + C

=12

ln |x2 + 4 | + C

2. ∫x

x2 + 4d x


Observing that dd x

(x2 + 4) = 2x:

∫x

x2 + 4d x =

12 ∫ 2x(x2 + 4)− 1

2 d x

=12

⋅(x2 + 4)1

2

12

+ C, by the reverse chain rule.

= x2 + 4 + C

Method 2: Algebraic Substitution #1

Observing that dd x

(x2 + 4) = 2x, u = x2 + 4 is a suitable choice of substitution:

∫x

x2 + 4d x = ∫ u− 1

2 ( 12

du) =

12 ∫ u− 1

2 du

= u12 + C

= x2 + 4 + C

Section 1

25

let u = x2 + 4

dud x

= 2x

x d x =12

du

let u = x2 + 4

dud x

= 2x

x d x =12

du

Questions 1 - 10


The goal of substitution is to eliminate difficult parts of the integrand, thus subbing u = x2 + 4 will also be suitable:

∫x

x2 + 4d x = ∫

uu

du

= ∫ du

= u + C

= x2 + 4 + C

Method 4: Trigonometric Substitution

We can make use of the identity tan2 θ + 1 = sec2 θ to help simplify the integrand:

∫x

x2 + 4d x = ∫

2 tan θ

4 tan2 θ + 4(2 sec2 θ) dθ

= ∫2 tan θ2 sec θ

(2 sec2 θ) dθ

= 2∫ tan θ sec θ dθ

= − 2∫ (cos θ)−2 (−sin θ) dθ

=2(cos θ)−1

−1× (−1) + C

= 2 sec θ + C

We then substitute back in:

= x2 + 4 + C

26

let u = x2 + 4

dud x

=x

x2 + 4

x d x = u du

let x = 2 tan θ

d xdθ

= 2 sec2 θ

d x = 2 sec2 θ dθ

θx

2

x2 + 4tan θ =

x2

⟹ sec θ =x2 + 4

2

3. ∫5x + 2x2 − 4

d x

Method 1: Partial Fractions

Observing that the denominator has two linear factors, we shall use partial fraction decomposition.

Let 5x + 2x2 − 4

=A

x + 2+

Bx − 2

.

∴ 5x + 2 = A(x − 2) + B(x + 2)

letting x = 2: 12 = 4B ⟹ B = 3.

letting x = − 2: −8 = − 4A ⟹ A = 2.

Hence, ∫5x + 2x2 − 4

d x = ∫ ( 2x + 2

+3

x − 2 ) d x

= 2 ln |x + 2 | + 3 ln |x − 2 | + C

4. ∫ sin x cos3 x d x


Observing that dd x

(cos x) = − sin x:

∫ sin x cos3 x d x = − ∫ (−sin x) cos3 x d x

= −cos4 x

4+ C, by the reverse chain rule.

Method 2: Substitution

Observing that dd x

(cos x) = − sin x, u = cos x is a suitable substitution:

∫ sin x cos3 x d x = ∫ u3(−du)

= − ∫ u3 du

= −u4

4+ C

= −cos4 x

4+ C

27

let u = cos x

dud x

= − sin x

sin x d x = − du

5. ∫ sin x sec3 x d x


Sine and cosine are very related functions, so we rewrite sec x in terms of cos x.

∫ sin x sec3 x d x = ∫ sin x (cos x)−3 d x

Observing that dd x

(cos x) = − sin x

= − ∫ (−sin x)(cos x)−3 d x

= −(cos x)−2

−2

=12

sec2 x + C


Note that sec x =1

cos x, and that d

d x(cos x) = − sin x, hence

u = cos x will be a suitable substitution:

∫ sin x sec3 x d x = ∫ u−3 (−du)

= −u−2

−2+ C

Substituting back in:

=12

sec2 x + C

6. ∫ cos2 x2

d x

Method 1: Half Angle Identity

Dealing with powers of trigonometric functions is not too nice, so we will use the half-angle identity to help turn the integrand into a single power:

As cos2 θ =12

(1 + cos 2θ):

∫ cos2 x2

d x =12 ∫ (1 + cos x) d x

=12

(x + sin x) + C

=x2

+sin x

2+ C

28

let u = cos x

dud x

= − sin x

sin x d x = − du

7. ∫ x sin x d x

Method 1: Integration by Parts

Notice that the integrand is a product of two functions, this hints that integration by parts will help.

Letting:

u x

sin x1

−cos x

u′�

v

v′�

∫ x sin x d x = − x cos x + ∫ cos x d x

= − x cos x + sin x + C

8. ∫ x sec2 2x d x


Notice that the integrand is a product of two functions, this hints that integration by parts will help.

Letting:

∫ x sec2 2x d x =x2

tan 2x −12 ∫ tan 2x d x

=x2

tan 2x −12 ∫

sin 2xcos 2x

d x

We then note that dd x

(cos 2x) = − 2 sin 2x and we manipulate the

integral such that we can apply the reverse chain rule:

=x2

tan 2x +14 ∫

(−2 sin 2x)cos 2x

d x

=x2

tan 2x +14

ln |cos 2x | + C

29

u x

sec2 2x1

12

tan 2x

u′�

v

v′�

9. ∫ tan−1 2x d x


This integral is a type of integral which doesn’t succumb to normal integration techniques, thus we shall try integration by parts, where v′� = 1.

Letting:

∫ tan−1 2x d x = x tan−1 2x − ∫2x

1 + 4x2d x

= x tan−1 2x −14 ∫

8x1 + 4x2

d x

= x tan−1 2x −14

ln |1 + 4x2 | + C

10. ∫x3

x2 + 1d x

Method 1: Long Division

We observe the degree of the numerator is greater than the denominator, thus the best course of action is polynomial long division. This yields:

x3

x2 + 1= x −

xx2 + 1

Hence, ∫x3

x2 + 1d x = ∫ x d x −

12 ∫

2xx2 + 1

d x

=x2

2−

12

ln |x2 + 1 | + C

Method 2: Algebraic Manipulation

Our goal with the manipulation is to obtain x2 + 1 on the numerator so that we are able to simplify the integral:

∫x3

x2 + 1d x = ∫

x3 + x − xx2 + 1

d x

= ∫x(x2 + 1)

x2 + 1d x − ∫

xx2 + 1

= ∫ x d x −12 ∫

2xx2 + 1

d x

=x2

2−

12

ln |x2 + 1 | + C

u tan−1 2x

12

1 + 4x2

x

u′�

v

v′�

30

11. ∫x

(x + 2)(x + 4)d x


Observing that the denominator has two linear factors, we shall use partial fraction decomposition.

Let x(x + 2)(x + 4)

=A

x + 2+

Bx + 4

∴ x = A(x + 4) + B(x + 2)

letting x = − 2, 2A = − 2 ⟹ A = − 1.

letting x = − 4, −2B = − 4 ⟹ B = 2

∴x

(x + 2)(x + 4)=

−1x + 2

+2

x + 4

∫x

(x + 2)(x + 4)d x = ∫ ( 2

x + 4−

1x + 2 ) d x

= 2 ln |x + 4 | − ln |x + 2 | + C

12. ∫(x − 1)(x + 1)(x − 2)(x − 3)

d x


The degree of the numerator is the same as the denominator, so we first need to long divide. After that, our new fraction has linear factors in the denominator, so we apply partial fractions.

∫(x − 1)(x + 1)(x − 2)(x − 3)

d x = ∫x2 − 1

x2 − 5x + 6d x

Now, x2 − 1x2 − 5x + 6

= 1 +5x − 7

x2 − 5x + 6 , by long division.

Using partial fractions, let 5x − 7(x − 2)(x − 3)

=A

x − 2+

Bx − 3

.

∴ A(x − 3) + B(x − 2) = 5x − 7

letting x = 3, B = 15 − 7 ⟹ B = 8.

letting x = 2, −A = 3 ⟹ A = − 3

∴ ∫(x − 1)(x + 1)(x − 2)(x − 3)

d x = ∫ 1 d x + 8∫1

x − 3d x − 3∫

1x − 2

d x

= x + 8 ln |x − 3 | − 3 ln |x − 2 | + C

Section 2

Questions 11 - 20

31

13. ∫2x − 1

x2 + 2x + 3d x


Observing that dd x

(x2 + 2x + 3) = 2x + 2, we shall split the

numerator:

∫2x − 1

x2 + 2x + 3d x = ∫ ( 2x + 2

x2 + 2x + 3−

32 + (x + 1)2 ) d x

We then apply the reverse chain rule to the first fraction, and the standard integral result to the second fraction:

= ln |x2 + 2x + 3 | −3

2tan−1 ( x + 1

2 ) + C

14. ∫x3

2x − 1d x

Method 1: Long Division

Noticing that the degree of the numerator is greater than the degree of the denominator, we shall use long division.

x3

2x − 1=

12

x2 +14

x +18

+18

2x − 1, by long division.

∴ ∫x3

2x − 1d x = ∫ [ 1

2x2 +

14

x +18

+1

8(2x − 1) ] d x

= ∫ ( 12

x2 +14

x +18 ) d x +

116 ∫

22x − 1

d x

=x3

6+

x2

8+

x8

+116

ln |2x − 1 | + C

32

15. ∫1 + x

1 − x − x2d x


Observing that dd x

(1 − x − x2) = − 1 − 2x, we can split the

numerator as 1 + x = −12

(−2x − 1) +12

:

∫1 + x

1 − x − x2d x = −

12 ∫

−2x − 1

1 − x − x2d x +

12 ∫

1

1 − x − x2d x

For our remaining integral, we then complete the square.

= −12 ∫ (−1 − 2x)(1 − x − x2)− 1

2 d x + 12 ∫

1

54 − (x + 1

2 )2

d x

These integrals are now all standard integrals:

= −12

(1 − x − x2)12

12

+12

sin−1x + 1

2

52

+ C

= − 1 − x − x2 +12

sin−1 ( 2x + 1

5 ) + C

16. ∫d x

x2(1 − x2)12


We can use the Pythagorean Identity of 1 − sin2 θ = cos2 θ to help simplify the integral:

∫d x

x2(1 − x2)12

= ∫cos θ

sin2 θ 1 − sin2 θdθ

= ∫1

sin2 θdθ

= ∫ cosec2θ dθ

= − cot θ + C

= −cos θsin θ

+ C


= −1 − x2

x+ C

33

let x = sin θ

d xdθ

= cos θ

d x = cos θ dθ

θx

1 − x2

1sin θ =

x1

⟹ cos θ = 1 − x2

17. ∫d x

x a2 + x2


We can use the Pythagorean Identity of 1 + tan2 θ = sec2 θ to help simplify the integral:

∫d x

x a2 + x2= ∫

a sec2 θ

a tan θ a2 + a2 tan2 θdθ

= ∫sec θ

a tan θdθ

=1a ∫ cosec θ dθ

= −1a

ln |cosec θ + cot θ | + C

By drawing a triangle, we can find cosec θ and cot θ:

θx

a

x2 + a2xa

= tan θ

= −1a

lnax

+a2 + x2

x+ C

18. ∫d x

x a2 − x2


We can use the Pythagorean Identity of 1 − sin2 θ = cos2 θ to help simplify the integral:

∫d x

x a2 − x2= ∫

a cos θ

a sin θ a2 − a2 sin2 θdθ

=1a ∫ cosec θ dθ

= −1a

ln |cosec θ + cot θ | + C

By drawing a triangle, we can find cosec θ and cot θ:

θx

a2 − x2

axa

= sin θ

= −1a

lnax

+a2 − x2

x+ C

34

let x = a tan θ

d xdθ

= a sec2 θ

d x = a sec2 θ dθ

let x = a sin θ d x = a cos θ dθ

a

19. ∫d x

x x2 − a2


We can use the Pythagorean Identity of sec2 θ − 1 = tan2 θ to help simplify the integral:

∫d x

x x2 − a2= ∫

a sec θ tan θ

a sec θ a2 sec2 θ − a2dθ

=1a ∫ dθ

=1a

θ + C

Now x = a sec θ ⟹ θ = sec−1 xa

:

=1a

sec−1 xa

+ C

20. ∫x

x + 1d x


We can use the substitution u = x to simplify the integrand.

∫x

x + 1d x = 2∫

u3

u + 1du

Now, u3

u + 1= u2 − u + 1 +

−1u + 1

, by long division

∴ ∫x

x + 1d x = 2∫ (u2 − u + 1 −

1u + 1 ) du

= 2 [ u3

3−

u2

2+ u − ln |u + 1 |] + C

Now substituting back in:

=23

x32 − x + 2x

12 − 2 ln ( x + 1) + C

35

let x = a sec θ d x = a sec θ dθ

let u = x d x = 2u du

21. ∫cos−1 x

1 − x2d x


∫cos−1 x

1 − x2d x = − ∫ u du

= −u2

2+ C


= −(cos−1 x)2

2+ C


Observe that dd x

(cos−1 x) = −1

1 − x2. Hence:

∫cos−1 x

1 − x2d x = − ∫ cos−1 x (−

1

1 − x2 ) d x

= −(cos−1 x)2

2+ C

22. ∫x + 1x − 1

d x (no longer on standard integrals)

Method 1: Rationalising the Numerator

We rationalise the numerator form a quadratic on the denominator:

∫x + 1x − 1

d x = ∫x + 1x − 1

×x + 1

x + 1d x

= ∫x + 1

x2 − 1d x

= ∫x

x2 − 1d x + ∫

1

x2 − 1d x

The rightmost integral used to be a standard integral, but has been removed from the formula sheet. Instead, we multiply top and bottom by x + x2 − 1.

= x2 − 1 + ∫( x + x2 − 1

x2 − 1 )x + x2 − 1

d x

Observe that the numerator is the derivative of the denominator:

´ = x2 − 1 + ln x + x2 − 1 + C

Section 3

36

let u = cos−1 x

dud x

= −1

1 − x2

Questions 21 - 30

23. ∫d x

x(ln x)3


Observing that dd x

(ln x) =1x

:

∫d x

x(ln x)3= ∫ (ln x)−3 ⋅

1x

d x

=(ln x)−2

−2+ C

= −1

2(ln x)2+ C


Observing that dd x

(ln x) =1x

, subbing u = ln x will be suitable:

∫d x

x(ln x)3= ∫ u−3 du

= −u−2

2+ C


= −1

2(ln x)2+ C

24. ∫ sec4 3x d x

Method 1: Trigonometric Identity and Reverse Chain Rule

We can use the Pythagorean Identity that sec2 θ = 1 + tan2 θ to make use of the reverse chain rule:

∫ sec4 3x d x = ∫ sec2 3x(1 + tan2 3x) d x

= ∫ sec2 3x d x + ∫ sec2 3x tan2 3x d x

The leftmost integral is now a standard result, so we now manipulate the remaining integral. Noting that d

d x(tan 3x) = 3 sec2 3x, we can apply the reverse chain rule:

= ∫ sec2 3x d x +13 ∫ (tan 3x)2(3 sec2 3x) d x

=tan 3x

3+

(tan 3x)3

9+ C

37

let u = ln x

dud x

=1x

d xx

= du

25. ∫d x

x2(1 − x)


Observing that the denominator consists of several factors, we can use partial fraction decomposition.

Let 1x2(1 − x)

=Ax

+Bx2

+C

1 − x.

∴ Ax(1 − x) + B(1 − x) + Cx2 = 1

letting x = 1, C = 1

letting x = 0, B = 1

letting x = − 1, −2A + 2B + C = 1 ⟹ A = 1

∴ ∫d x

x2(1 − x)= ∫ ( 1

x+

1x2

+1

1 − x ) d x

= ln |x | −1x

− ln |1 − x | + C

= lnx

1 − x−

1x

+ C

26. ∫d x

x2(1 + x2)


We can use the Pythagorean Identity 1 + tan2 θ = sec2 θ to help simplify the integrand.

∫d x

x2(1 + x2)= ∫

sec2 θtan2 θ(1 + tan2 θ)

dθ

= ∫ cot2 θ dθ

= ∫ (cosec2 θ − 1) dθ

= ∫ cosec2θ dθ − ∫ dθ

= − cot θ − θ + C


= −1x

− tan−1 x + C

38

let x = tan θ

d xdθ

= sec2 θ

d x = sec2 θ dθ

27. ∫d x

(1 + x2)2



∫d x

(1 + x2)2= ∫

sec2 θ(1 + tan2 θ)2

dθ

= ∫sec2 θsec4 θ

dθ

= ∫ cos2 θ dθ

=12 ∫ (1 + cos 2θ) dθ

=12 (θ +

sin 2θ2 ) + C

=12

(θ + cos θ sin θ) + C

By drawing a triangle, we can find cos θ and sin θ:

=tan−1 x

2+

x2(x2 + 1)

+ C

28. ∫ tan3 x d x

Method 1: Trigonometric Identity


∫ tan3 x d x = ∫ tan x(sec2 x − 1) d x

= ∫ tan x sec2 x d x − ∫ tan x d x

= ∫ tan x sec2 x d x − ∫sin xcos x

d x

= ∫ tan x sec2 x d x + ∫(−sin x)

cos xd x

=tan2 x

2+ ln |cos x | + C, by the reverse chain rule.

39

let x = tan θ

d xdθ

= sec2 θ

d x = sec2 θ dθ

θx

1

x2 + 1

x1

= tan θ

⟹ sin θ =x

x2 + 1, cos θ =

1

x2 + 1

29. ∫d x

5 + 3 cos x

Method 1: t-Substitution

There are no obvious/nice manipulations such substitutions, so we will apply the t-substitution to convert the trigonometric terms into algebraic terms:

∫d x

5 + 3 cos x= ∫

1

5 + 3(1 − t2)1 + t2

⋅2

1 + t2dt

= ∫1 + t2

5 + 5t2 + 3 − 3t2⋅

21 + t2

dt

= 2∫dt

2t2 + 8

= ∫dt

t2 + 4

=12

tan−1 t2

+ C

Subbing back in:

=12

tan−1 (tan x

2

2 ) + C

30. ∫d x

3 + 5 cos x


There are no obvious/nice manipulations such substitutions, so we will apply the t-substitution to convert the trigonometric terms into algebraic terms:

∫d x

3 + 5 cos x= ∫

1

3 + 5 − 5t2

1 + t2

⋅2

1 + t2dt

= ∫1 + t2

3 + 3t2 + 5 − 5t2⋅

21 + t2

dt

= ∫2

8 − 2t2dt

= ∫1

4 − t2dt

From here we need to decompose the integrand into partial

fractions. Let 14 − t2

=A

2 + t+

B2 − t

.

∴ A(2 − t) + B(2 + t) = 1

letting t = 2, 4B = 1, B =14

letting t = − 2, 4A = 1, A =14

40

let t = tanx2

d x =2

1 + t2dt

cos x =1 − t2

1 + t2

let t = tanx2

d x =2

1 + t2dt

cos x =1 − t2

1 + t2

=14 ∫ ( 1

2 + t+

12 − t ) dt

=14 (ln |2 + t | − ln |2 − t |) + C

=14

ln2 + t2 − t

+ C

Now subbing back in:

=14

ln2 + tan x

2

2 − tan x2

+ C

41

31. ∫sin x

5 + 3 cos xd x


Observing that dd x

(5 + 3 cos x) = − 3 sin x:

∫sin x

5 + 3 cos xd x = −

13 ∫

−3 sin x5 + 3 cos x

d x

= −13

ln |5 + 3 cos x | + C, by the reverse chain rule.


u = cos x, u = 3 cos x, u = 5 + 3 cos x are all effective substitutions.

However, u = 5 + 3 cos x will simplify the integrand the most:

∫sin x

5 + 3 cos xd x = ∫

1u (−

13

du) = −

13

ln |u | + C

= −13

ln |5 + 3 cos x | + C

32. ∫d x

1 + cos2 x


There are no nice substitutions, so we apply the t-substitution, but before we do so, we need to remove any powers in order to simplify the integrand:

Applying the double result cos2 x =12

(1 + cos 2x):

∫d x

1 + cos2 x= ∫

d x

1 + cos 2x2 + 1

2

= ∫2

cos 2x + 3d x

= ∫2

1 − t2

1 + t2 + 3×

dt1 + t2

Section 4

42

let t = tan x

d x =1

1 + t2dt

cos 2x =1 − t2

1 + t2

let u = 5 + 3 cos x

dud x

= − 3 sin x

sin x d x = −13

du

Questions 31 - 40

= ∫2

1 − t2 + 3 + 3t2

1 + t2

×dt

1 + t2

= ∫2

2t2 + 4dt

= ∫1

t2 + 2dt

=1

2tan−1 ( t

2 ) + C

=1

2tan−1 ( tan x

2 ) + C


We divide the top and bottom by cos2 x:

∫d x

1 + cos2 x= ∫

sec2 xsec2 x + 1

d x

Applying the Pythagorean identity 1 + tan2 θ = sec2 θ:

= ∫sec2 x

2 + tan2 xd x

Noting that dd x

(tan x) = sec2 x, we can apply the reverse chain rule:

=1

2tan−1 ( tan x

2 ) + C

33. ∫d x

cos2 x2 − sin2 x

2

Method 1: Double Angle Identity

We observe that we can simplify the denominator using the double angle identity for cosine: cos 2θ = cos2 θ − sin2 θ:

∫d x

cos2 x2 − sin2 x

2

= ∫d x

cos x

= ∫ sec x d x

This is now an integral we have encountered before, and the trick is to multiply the top and bottom by sec x + tan x:

= ∫sec x(sec x + tan x)

sec x + tan xd x

= ∫sec x tan x + sec2 x

sec x + tan xd x

Observe that dd x

(sec x + tan x) = sec x tan x + sec2 x, thus we apply

the reverse chain rule:

= ln |sec x + tan x | + C

43

34. ∫ x2 sin x d x


The integrand is a product of two factors, so we integrate by parts.

Letting:

∫ x2 sin x d x = − x2 cos x + 2∫ x cos x d x

The next integral we have to deal with is another product of two factors, so we integrate by parts again.

Letting:

∫ x cos x d x = x sin x − ∫ sin x d x

= x sin x + cos x + C

Hence, ∫ x2 sin x d x = −x2 cos x + 2(x sin x + cos x) + C

= − x2 cos x + 2x sin x + 2 cos x + C

35. ∫x2

(x − 1)(x − 2)(x − 3)d x


Observing that the degree of the denominator is greater than the numerator, and that denominator has three linear factors, we shall use partial fraction decomposition.

Let x2

(x − 1)(x − 2)(x − 3)=

Ax − 1

+B

x − 2+

Cx − 3

∴ A(x − 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x − 2) = x2

letting x = 2: 0 − B + 0 = 4 ⟹ B = − 4

letting x = 1: 2A + 0 + 0 = 1 ⟹ A =12

letting x = 3: 0 + 0 + 2C = 9 ⟹ C =92

Hence, x2

(x − 1)(x − 2)(x − 3)=

12(x − 1)

−4

x − 2+

92(x − 3)

.

Thus:∫x2

(x − 1)(x − 2)(x − 3)d x

= ∫ ( 12(x − 1)

−4

x − 2+

92(x − 3) ) d x

=12

ln |x − 1 | − 4 ln |x − 2 | +92

ln |x − 3 | + C

44

u x2

sin x2x

−cos x

u′�

v

v′�

u x

cos x1

sin x

u′�

v

v′�

36. ∫ex

ex − 1d x


Observing that dd x

(ex − 1) = ex

∫ex

ex − 1d x = ln |ex − 1 | + C, by the reverse chain rule.


Observe that dd x

(ex − 1) = ex, so u = ex − 1 will be a suitable

substitution:

∫ex

ex − 1d x = ∫

duu

= ln |u | + C

= ln |ex − 1 | + C

37. ∫d x

3 sin2 x + 5 cos2 x


There are no “nice substitutions”, so we will use the t-substitution. Before we do so, we need to remove all powers from the integrand. To do so, we need to manipulate before we apply the double angle identity.

∫d x

3 sin2 x + 5 cos2 x= ∫

d x3(1 − cos2 x) + 5 cos2 x

= ∫d x

3 + 2 cos2 x

= ∫d x

3 + 2 ( cos 2x + 12 )

= ∫d x

cos 2x + 4

We are now able to carry out the t-substitution:

= ∫1

1 − t2

1 + t2 + 4×

dt1 + t2

= ∫1

1 − t2 + 4 + 4t2

1 + t2

×dt

1 + t2

= ∫1 + t2

3t2 + 5×

dt1 + t2

45

let t = tan x

d x =1

1 + t2dt

cos 2x =1 − t2

1 + t2

let u = ex − 1

dud x

= ex

ex d x = du

= ∫1

3t2 + 5dt

=13 ∫

1

t2 + 53

dt

=13 (

3

5 ) tan−1 35

t + C


=15

15tan−1 ( 3

5tan x) + C


We divide the top and bottom by cos2 x:

∫d x

3 sin2 x + 5 cos2 x= ∫

sec2 x3 tan2 x + 5

d x

Noting that dd x ( 3 tan x) = 3 sec2 x, we manipulate such that

we can apply the reverse chain rule:

=1

3 ∫3 sec2 x

( 3 tan x)2

+ 5d x

=1

15tan−1 ( 3

5tan x) + C

38. ∫ x3e5x4−7 d x


Observe that dd x

(5x4 − 7) = 20x3, thus we manipulate such that

we can apply the reverse chain rule:

∫ x3e5x4−7 d x =120 ∫ (20x3)e5x4−7 d x

=120

e5x4−7 + C


Observing that dd x

(5x4 − 7) = 20x3, u = 5x4 − 7 will be a suitable

substitution:

∫ x3e5x4−7 d x = ∫ eu ( 120

du) =

120

eu + C


=120

e5x4−7 + C

46

let u = 5x4 − 7

dud x

= 20x3

x3 d x =120

du

39. ∫ x5 ln x d x


Our integrand is a product of two functions, hence we integrate by parts:

Letting:

∫ x5 ln x d x =x6

6ln x −

16 ∫ x5 d x

=x6

6ln x −

16

⋅x6

6+ C

=x6

6ln x −

x6

36+ C

40. ∫3x + 2

x(x + 1)3d x


Observing that the degree of the denominator is greater than the numerator can contains linear factors, we decompose the integrand into partial fractions.

Let 3x + 2x(x + 1)3

=Ax

+B

(x + 1)+

C(x + 1)2

+D

(x + 1)3

∴ A(x + 1)3 + Bx(x + 1)2 + Cx(x + 1) + Dx = 3x + 2

letting x = 0: A + 0 + 0 + 0 = 2 ⟹ A = 2.

letting x = − 1: 0 + 0 + 0 − D = − 1 ⟹ D = 1.

letting x = 1: 8A + 4B + 2C + D = 5

Substituting our previous values: 4B + 2C = − 12

letting x = 2: 27A + 18B + 6C + 2D = 8

Substituting our previous values: 18B + 6C = − 48

We then solve:

2B + C = − 6 and 3B + C = − 8 simultaneously.

Subtracting these equations:

(3B + C ) − (2B + C ) = − 8 − (−6) ⟹ B = − 2

Substituting this back in: C = − 2.47

u ln x

x51x

x6

6

u′�

v

v′�

Hence: 3x + 2x(x + 1)3

=2x

−2

x + 1−

2(x + 1)2

+1

(x + 1)3

Thus:

∫3x + 2

x(x + 1)3d x = ∫ ( 2

x−

2x + 1

−2

(x + 1)2+

1(x + 1)3 ) d x

= 2 ln |x | − 2 ln |x + 1 | +2

x + 1−

12(x + 1)2

+ C

48

41. ∫ ln(x3) d x


We can simplify the integral using the log law: loga(xb) = b loga(x):

∫ ln(x3) d x = 3∫ ln x d x

We then compute the integral using integration by parts.

Letting:

= 3 (x ln x − ∫ x ⋅1x

d x)

= 3 (x ln x − x) + C

= 3x ln x − 3x + C

42. ∫d x

ex + e−x

Method 1: Algebraic Manipulation and Reverse Chain Rule

We multiply the top and bottom by ex:

∫d x

ex + e−x= ∫

ex

(ex)2 + 1d x

Observing dd x

(ex) = ex, this integral is now in a standard result:

= tan−1(ex) + C

Method 2: Algebraic Manipulation and Substitution

After the manipulation, instead of applying the reverse chain rule, we could instead apply the substitution u = ex.

∫d x

ex + e−x= ∫

ex

(ex)2 + 1d x

= ∫du

u2 + 1

= tan−1 u + C


= tan−1(ex) + c

Section 5

49

u ln x

11x

x

u′�

v

v′�

let u = ex

dud x

= ex

ex d x = du

Questions 41 - 50

43. ∫ (5x3 + 7x − 1)32 ⋅ (15x2 + 7) d x


Observe that dd x

(5x3 + 7x − 1) = 15x2 + 7, hence we can apply the

reverse chain rule directly.

∫ (5x3 + 7x − 1)32 ⋅ (15x2 + 7) d x =

(5x3 + 7x − 1)52

52

+ C

=25

(5x3 + 7x − 1)52 + C


Observing that dd x

(5x3 + 7x − 1) = 15x2 + 7, we let

u = 5x3 + 7x − 1 as our substitution.

∫ (5x3 + 7x − 1)32 ⋅ (15x2 + 7) d x = ∫ u

32 du

=u

52

52

+ C

=25

u52 + C


=25

(5x3 + 7x − 1)52 + C

44. ∫d x

(x2 + 1)(x2 + 4)

Method 1: Partial Fractions (faster method)

The denominator consists of two quadratic factors, so we decompose the integrand into partial fractions. While the method of substitution appears to be quite slow, we can use complex numbers to speed up the process.

Let 1(x2 + 1)(x2 + 4)

=Ax + Bx2 + 1

+Cx + Dx2 + 4

∴ (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1) = 1

letting x = i: (Ai + B)(−1 + 4) + 0 = 1 ⟹ Ai + B =13

Equating real and imaginary parts: B =13

, A = 0

letting x = 2i: 0 + (2Ci + D)(−4 + 1) = 1 ⟹ 2Ci + D = −13

Equating real and imaginary parts: D = −13

, C = 0

Hence: 1(x2 + 1)(x2 + 4)

=1

3(x2 + 1)−

13(x2 + 4)

.

Thus ∫d x

(x2 + 1)(x2 + 4)=

13 ∫

d xx2 + 1

−13 ∫

d xx2 + 4

=13

tan−1 x −16

tan−1 ( x2 ) + C

50

let u = 5x3 + 7x − 1

dud x

= 15x2 + 7

du = (15x2 + 7) d x


We decompose the integrand into partial fractions:

Let 1(x2 + 1)(x2 + 4)

=Ax + Bx2 + 1

+Cx + Dx2 + 4

∴ (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1) = 1

For non-4U students, we can still solve for the constants by either equating coefficients/substituting real values. Here we choose the method of substitution.

letting x = 0: 4B + D = 1 (1)

letting x = 1: 5A + 5B + 2C + 2D = 1 (2)

letting x = − 1: −5A + 5B − 2C + 2D = 1 (3)

letting x = 2: 16A + 8B + 10C + 5D = 1 (4)

We then solve simultaneously:

Adding equations (2) and (3): 10B + 4D = 2 (5)

Performing (5) −4 × (1): −6B = − 2 ⟹ B =13

.

Subbing B =13

back into (1): 43

+ D = 1 ⟹ D = −13

.

If we then sub these values of B and D into (2) and (4):

5A + 2C = 0 and 16A + 10C = 0

Solving these two new equations, we get that A = C = 0.

Hence: 1(x2 + 1)(x2 + 4)

=1

3(x2 + 1)−

13(x2 + 4)

.

Thus ∫d x

(x2 + 1)(x2 + 4)=

13 ∫

d xx2 + 1

−13 ∫

d xx2 + 4

=13

tan−1 x −16

tan−1 ( x2 ) + C

51

45. ∫ (x2 + x + 1)−1 d x

Method 1: Completing the Square

Observe that we have an irreducible quadratic in the denominator, so we complete the square and then apply our standard integral result.

∫ (x2 + x + 1)−1 d x = ∫d x

x2 + x + 1

= ∫d x

x + x + 12

2+ 1 − 1

2

2

= ∫1

(x + 12 )

2+ ( 3

2 )2 d x

=2

3tan−1

x + 12

32

+ C

=2

3tan−1 ( 2x + 1

3 ) + C

46. ∫ ex sin 2x d x


Our integrand is a product of two functions, so we apply integration by parts.

Letting:

∫ ex sin 2x d x = ex sin(2x) − 2∫ cos(2x)ex d x

Our new integral is another product of two functions, so we apply integration by parts again:

Letting:

= ex sin(2x) − 2 [ex cos(2x) + 2∫ ex sin(2x)] d x

= ex sin(2x) − 2ex cos(2x) − 4∫ ex sin(2x) d x

Notice that we have reformed our original integral. Rearranging:

5∫ ex sin(2x) = ex sin(2x) − 2ex cos(2x) + C1

Thus, ∫ ex sin 2x =ex

5 (sin 2x − 2 cos 2x) + C2

52

u sin(2x)

ex2 cos(2x)

ex

u′�

v

v′�

u cos(2x)

ex−2 sin(2x)

ex

u′�

v

v′�

47. ∫ (x2 + x − 1)−1 d x


After some algebraic manipulations, the quadratic in our denominator is a product of two linear factors, so we decompose the integrand into partial fractions.

∫ (x2 + x − 1)−1 d x = ∫1

(x + 12 )

2− 5

4

d x

= ∫d x

(x + 12 +

52 ) (x + 1

2 −5

2 )We now decompose into partial fractions.

Let 1x2 + x − 1

=A

x + 12 +

52

+B

x + 12 −

52

∴ A (x +12

−5

2 ) + B (x +12

+5

2 ) = 1

letting x = −12

+5

2: 5B = 1 ⟹ B =

1

5

letting x = −12

−5

2: − 5A = 1 ⟹ A = −

1

5

Hence: 1x2 + x − 1

= −

1

5

x + 12 +

52

+

1

5

x + 12 −

52

.

Thus:

∫ (x2 + x − 1)−1 d x = −1

5 ∫d x

x + 12 +

52

+1

5 ∫d x

x + 12 −

52

= −1

5ln x +

12

+5

2+

1

5ln x +

12

−5

2+ C

=1

5ln

x + 12 −

52

x + 12 +

52

+ C

=1

5ln

2x + 1 − 5

2x + 1 + 5+ C

53

48. ∫ (x2 − x)− 12 d x (no longer on standard integrals)

Method 1: Completing the Square

We complete the square for the quadratic in the denominator

∫ (x2 − x)− 12 d x = ∫

1

x2 − xd x

= ∫1

(x − 1)2 − 14

d x

This integral is now in the form of ∫d x

x2 − a2, which was

previously a standard integral but is now removed with the addition of the formula sheet.

= ln x −12

+ x2 − x + C1

= ln 2x − 1 + 2 x2 − x + C2

49. ∫1 − 2x3 + x

d x


Our goal is to manipulate the numerator into a multiple of the denominator.

1 − 2x = 1 − 2(x + 3 − 3)

= 7 − 2(x + 3)

Hence: ∫1 − 2x3 + x

d x = ∫7 − 2(x + 3)

3 + xd x

= ∫7

3 + xd x − 2∫ d x

= 7 ln |x + 3 | − 2x + C

54

50. ∫ x3(4 + x2)− 12 d x


Although a slightly less obvious substitution, observe that d

d x(4 + x2) = 2x, and that x3 = x × x2. Hence we choose

u = 4 + x2 as our substitution.

∫ x3(4 + x2)− 12 d x = ∫ (u − 4)u− 1

2 ( 12

du) =

12 ∫ (u

12 − 4u− 1

2 ) du

=12 ( 2

3u

32 − 8u

12) + C


=13

(x2 + 4)32 − 4(x2 + 4)1

2 + C

=x2 + 4

3 (x2 + 4 − 12) + C

=13

(x2 − 8) x2 + 4 + C


Another suitable substitution is u = x2 + 4. Notice that this removes the “most difficult” part of the integrand, which is the radical in the denominator.

∫ x3(4 + x2)− 12 d x = ∫

x3

4 + x2d x

= ∫x

4 + x2⋅ x2 d x

= ∫ (u2 − 4) du

=u3

3− 4u + C

=13

u(u2 − 12) + C


=13

x2 + 4(x2 + 4 − 12) + C

=13

(x2 − 8) x2 + 4 + C

55

let u = x2 + 4

⟺ x2 = u − 4

dud x

= 2x

x d x =12

du

let u = x2 + 4

⟺ x2 = u2 − 4

dud x

=x

x2 + 4

du =x

x2 + 4d x


Another potential substitution is a trigonometric substitution, utilising the identity 1 + tan2 θ = sec2 θ. Although not the most efficient, it is still a perfectly valid method.

∫ x3(4 + x2)− 12 d x = ∫

x3

4 + x2d x

= ∫8 tan3 θ

4 + 4 tan2 θ(2 sec2 θ) dθ

= ∫16 tan3 θ sec2 θ

2 sec θdθ

= 8∫ tan3 θ sec θ dθ

= 8∫sin3 θcos4 θ

dθ

= 8∫sin θ(1 − cos2 θ)

cos4 θdθ

= 8∫sin θ

cos4 θdθ − 8∫

sin θcos2 θ

dθ

= −83

(cos θ)−3 + 8(cos θ)−1 + C

=8

cos θ−

83 cos3 θ

+ C

We then use a triangle to find the value of cos θ to substitute back in:

Hence:

∫ x3(4 + x2)− 12 d x = 8 ⋅

x2 + 42

−83

⋅(x2 + 4) x2 + 4

8+ C

= 4 x2 + 4 −13

(x2 + 4) x2 + 4 + C

=13

(x2 − 8) x2 + 4 + C

56

let x = 2 tan θ

d xdθ

= 2 sec2 θ

d x = 2 sec2 θ dθ

θx

2

x2 + 4tan θ =x2

⟹ cos θ =2

x2 + 4

51. ∫sin 2x

3 cos2 x + 4 sin2 xd x

Method 1: Manipulation and Reverse Chain Rule

Observe that dd x

(sin2 x) = 2 sin x cos x = sin 2x, and that we can

change cos2 x = 1 − sin2 x.

∫sin 2x

3 cos2 x + 4 sin2 xd x = ∫

2 sin x cos x3(1 − sin2 x) + 4 sin2 x

d x

= ∫2 sin x cos x3 + sin2 x

d x

= ln |3 + sin2 x | + C, by the reverse chain rule.

Now as 3 + sin2 x > 0 for all real x, we can remove the absolute values:

= ln(3 + sin2 x) + C

Method 2: Manipulation and Substitution

Observing that dd x

(sin2 x) = 2 sin x cos x = sin 2x, we can apply the

same manipulation as in Method 1, but instead apply the substitution u = sin2 x.

∫sin 2x

3 cos2 x + 4 sin2 xd x = ∫

2 sin x cos x3(1 − sin2 x) + 4 sin2 x

d x

= ∫2 sin x cos x3 + sin2 x

d x

= ∫du

3 + u

= ln |3 + u | + C


= ln |3 + sin2 x | + C

Now as 3 + sin2 x > 0 for all real x, we can remove the absolute values:

= ln(3 + sin2 x) + C

Section 6

57

Questions 51 - 60

52. ∫x2

1 − x4d x


We observe that the denominator can be factorised into linear and quadratic factors, which we can then decompose into partial fractions.

∫x2

1 − x4d x = ∫

x2

(1 − x2)(1 + x2)d x

= ∫x2

(1 − x)(1 + x)(1 + x2)d x

Let x2

(1 − x2)(1 + x2)=

A1 − x

+B

1 + x+

Cx + Dx2 + 1

∴ x2 = A(1 + x)(x2 + 1) + B(1 − x)(x2 + 1) + (Cx + D)(1 − x2)

letting x = i: −1 = 0 + 0 + 2Ci + 2D

Equating real and imaginary parts: D = −12

, C = 0.

letting x = 1: 1 = 4A ⟹ A =14

letting x = − 1: 1 = 0 + 4B ⟹ B =14

Hence: x2

(1 − x2)(1 + x2)=

14

1 − x+

14

1 + x−

12

x2 + 1.

Thus:

∫x2

1 − x4d x =

14 ∫

d x1 − x

+14 ∫

d x1 + x

−12 ∫

d xx2 + 1

=14

ln |1 + x | −14

ln |1 − x | −12

tan−1 x + C

58

53. ∫d x

sin x cos x

Method 1: Double Angle Formula

We use the double angle for sine: sin 2x = 2 sin x cos x to simplify the integral:

∫d x

sin x cos x= ∫

22 sin x cos x

d x

= ∫2 d xsin 2x

= 2∫ cosec x d x

The integral of cosec x is another difficult one, but the manipulation we use is to multiply the top and bottom by cosec x + cot x.

= 2∫(cosec x + cot x)cosec x

cosec x + cot xd x

Observe that dd x

(cosec x + cot x) = − cosec x cot x − cosec2x, so we

can apply the reverse chain rule.

= − 2 ln |cosec x + cot x | + C


If we divide the top and bottom by cos2 x:

∫d x

sin x cos x= ∫

sec2 xtan x

d x

Note that dd x

(tan x) = sec2 x, so we apply the reverse chain rule:

= ln | tan x | + C

59

54. ∫ ln x − 1 d x


We make use of the log law loga(xb) = b loga(x), and then much

like how we evaluate ∫ ln x d x, we integrate by parts.

∫ ln x − 1 d x =12 ∫ ln(x − 1) d x

Letting:

=12 (x ln(x − 1) − ∫

xx − 1

d x)To evaluate the new integral, we will use an algebraic manipulation:

=12

x ln(x − 1) −12 ∫

x − 1 + 1x − 1

d x

=12

x ln(x − 1) −12 ∫ (1 +

1x − 1 ) d x

=12

x ln(x − 1) −12

x −12

ln |x − 1 | + C

As x is restricted to x − 1 > 0 due to the domain of the integrand, we can omit the absolute value of the logarithm:

=12

(x − 1)ln(x − 1) −12

x + C

55. ∫d x

ex − 1


We divide the top and bottom of the integrand by ex:

∫d x

ex − 1= ∫

e−x

1 − e−xd x

Observe dd x

(1 − e−x) = e−x, thus we apply the reverse chain rule.

= ln |1 − e−x | + C

Method 2: Substitution and Partial Fractions

The best substitution here is u = ex, in order to eliminate the “most difficult” part of the integrand.

∫d x

ex − 1= ∫

1u(u − 1)

du

Our denominator now consists of two linear factors, so we decompose the integrand into partial fractions

Let 1u(u − 1)

=Au

+B

u − 1.

∴ A(u − 1) + Bu = 1

letting u = 1: B = 1 and letting u = 0: A = − 1

60

u ln(x − 1)

11

x − 1

x

u′�

v

v′�

let u = ex

dud x

= ex

d x =1u

du

Hence: 1u(u − 1)

= −1u

+1

u − 1.

Thus:

∫d x

ex − 1= ∫

duu − 1

− ∫duu

= ln |u − 1 | − ln |u | + C


= ln |ex − 1 | − ln |ex | + C

= ln |ex − 1 | − x + C

While this result appears different from the one obtained in Method 1, note that these two forms are equivalent when suitable log laws are applied.

56. ∫sec2 x

tan2 x − 3 tan x + 2d x


Note that dd x

(tan x) = sec2 x, thus we choose u = tan x as our

substitution.

∫sec2 x

tan2 x − 3 tan x + 2d x = ∫

duu2 − 3u + 2

= ∫du

(u − 2)(u − 1)

Our denominator now consists of two linear factors, so we decompose the integrand into partial fractions.

Let 1(u − 2)(u − 1)

=A

u − 2+

Bu − 1

.

∴ A(u − 1) + B(u − 2) = 1

letting u = 1: −B = 1 ⟹ B = − 1

letting u = 2: A = 1

Hence: 1(u − 2)(u − 1)

=1

u − 2−

1u − 1

.

Thus: ∫sec2 x

tan2 x − 3 tan x + 2d x = ∫

duu − 2

− ∫du

u − 1

= ln |u − 2 | − ln |u − 1 | + C


= ln | tan x − 2 | − ln | tan x − 1 | + C

61

57. ∫x + 1

(x2 − 3x + 2)12

d x (no longer on standard integrals)

Method 1: Algebraic Manipulation and Reverse Chain Rule

Our goal is to manipulate the numerator into a multiple of the

derivative of the denominator. As dd x

(x2 − 3x + 2) = 2x − 3, we

manipulate x + 1 as so: x + 1 =12

(2x − 3) +32

+ 1.

∫x + 1

(x2 − 3x + 2)12

d x = ∫12 (2x − 3) + 5

2

x2 − 3x + 2d x

=12 ∫

2x − 3

x2 − 3x + 2d x +

52 ∫

d x

x2 − 3x + 2

We apply the reverse chain rule to the leftmost integral, and complete the square for the rightmost integral.

= x2 − 3x + 2 +52 ∫

d x

(x − 32 )

2− 1

4

= x2 − 3x + 2 +52

ln x −32

+ x2 − 3x + 2 + C1

= x2 − 3x + 2 +52

ln 2x − 3 + 2 x2 − 3x + 2 + C2

58. ∫ sin 2x cos x d x

Method 1: Double Angle Identity

We apply the double angle formula for sine: sin 2x = 2 sin x cos x.

∫ sin 2x cos x d x = 2∫ sin x cos2 x d x

We now observe that dd x

(cos x) = − sin x, so we apply the reverse

chain rule: (alternatively you can make a substitution u = cos x)

= −23

cos3 x + C

Method 2: Product to Sums

Our integrand consists of a product of trigonometric functions, so we use the product to sum identity:

sin A cos B =12 [sin(A + B) + sin(A − B)], setting A = 2x and B = x.

∫ sin 2x cos x d x =12 ∫ (sin 3x + sin x) d x

= −16

cos 3x −12

cos x + C

62

59. ∫x

1 + x3d x


We can factorise the denominator into a quadratic and linear factor, allowing us to decompose the integrand into partial fractions.

∫x

1 + x3d x = ∫

x(1 + x)(1 − x + x2)

d x

Now, let x(1 + x)(1 − x + x2)

=A

1 + x+

Bx + C1 − x + x2

∴ A(1 − x + x2) + (Bx + C )(1 + x) = x

letting x = − 1: 3A = − 1 ⟹ A = −13

letting x = 0: A + C = 0 ⟹ C =13

letting x = 1: A + 2B + 2C = 1 ⟹ B =13

Hence: x(1 + x)(1 − x + x2)

= −1

3(1 + x)+

x + 13(1 − x + x2)

Thus: ∫x

1 + x3d x = −

13 ∫

d x1 + x

+13 ∫

x + 11 − x + x2

d x

In order to simplify the right integral, our goal is to express the numerator is a multiple of the derivative of the denominator.

We do so through: x + 1 =12

(2x − 1) +32

.

Hence:

∫x

1 + x3d x = −

13 ∫

d x1 + x

+16 ∫

2x − 11 − x + x2

d x +12 ∫

d x1 − x + x2

d x

We then complete the square for the right most integral.

= −13 ∫

11 + x

d x +16 ∫

2x − 11 − x + x2

d x +12 ∫

1

(x − 12 )

2+ 3

4

d x

= −13

ln |1 + x | +16

ln |1 − x + x2 | +12

⋅2

3tan−1

x − 12

32

+ C

= −13

ln |1 + x | +16

ln |1 − x + x2 | +1

3tan−1 ( 2x − 1

3 ) + C

63

60. ∫ x tan−1 x d x


Our integrand is a product of two functions, hence we apply integration by parts.

Letting:

∫ x tan−1 x d x =x2

2tan−1 x −

12 ∫

x2

1 + x2d x

=x2

2tan−1 x −

12 ∫ ( x2 + 1

1 + x2−

11 + x2 ) d x

=x2

2tan−1 x −

12 ∫ (1 −

11 + x2 ) d x

=x2

2tan−1 x −

12

x +12

tan−1 x + C

64

u tan−1 x

x1

1 + x2

12

x2

u′�

v

v′�

61. ∫ (1 + 3x + 2x2)−1 d x


Our denominator can be factorised into two linear factors, so we decompose the integrand into partial fractions.

∫ (1 + 3x + 2x2)−1 d x = ∫1

(2x + 1)(x + 1)d x

Let 1(2x + 1)(x + 1)

=A

2x + 1+

Bx + 1

.

∴ A(x + 1) + B(2x + 1) = 1

letting x = − 1: −B = 1 ⟹ B = − 1

letting x = −12

: 12

A = 1 ⟹ A = 2

Hence: 1(2x + 1)(x + 1)

=2

2x + 1−

1x + 1

Thus: ∫ (1 + 3x + 2x2)−1 d x = ∫2

2x + 1d x − ∫

d xx + 1

= ln |2x + 1 | − ln |x + 1 | + C

62. ∫ (9 − x2)12 d x


We make use of the Pythagorean identity 1 − sin2 θ = cos2 θ in order to simplify the integral.

∫ (9 − x2)12 d x = ∫ 9 − 9 sin2 θ(3 cos θ) dθ

= ∫ 9 cos2 θ(3 cos θ) dθ

= 9∫ cos2 θ dθ

=92 ∫ (1 + cos 2θ) dθ

=92 (θ +

12

sin 2θ) + C

=92

θ +92

sin θ cos θ + C

Section 7

Questions 61 - 70

65

let x = 3 sin θ

d xdθ

= 3 cos θ

d x = 3 cos θ dθ


=92

sin−1 x3

+92

⋅x3

⋅9 − x2

3+ C

=12 [9 sin−1 ( x

3 ) + x 9 − x2] + C

63. ∫ (9 + x2)12 d x


We can make use of the Pythagorean identity 1 + tan2 θ = sec2 θ to help simplify the integral.

∫ (9 + x2)12 d x = ∫ 9 + 9 tan2 θ(3 sec2 θ) dθ

= ∫ 9 sec2 θ(3 sec2 θ) dθ

= 9∫ sec3 θ dθ

This integral we evaluate using integration by parts.

Letting:

∫ sec3 θ dθ = sec θ tan θ − ∫ sec θ tan2 θ dθ

= sec θ tan θ − ∫ sec θ(sec2 θ − 1) dθ

= sec θ tan θ − ∫ sec3 θ dθ + ∫ sec θ dθ

2∫ sec3 θ dθ = sec θ tan θ + ln |sec θ + tan θ | + C

66

θx

9 − x2

3x3

= sin θ

⟹ cos θ =9 − x2

3

u sec θ

sec2 θsec θ tan θ

tan θ

u′�

v

v′�

let x = 3 tan θ

d xdθ

= 3 sec2 θ

d x = 3 sec2 θ dθ

Hence:

∫ (9 + x2)12 d x =

92

sec θ tan θ +92

ln |sec θ + tan θ | + C


=92

⋅x2 + 9

3⋅

x3

+92

lnx2 + 9

3+

x3

+ C1

=12

x x2 + 9 +92

lnx + x2 + 9

3+ C1

=12

x x2 + 9 +92

ln x + x2 + 9 + C2

64. ∫ x(9 + x2)12 d x


Observe that dd x

(x2 + 9) = 2x, hence we can apply the reverse

chain rule.

∫ x(9 + x2)12 d x =

12 ∫ (2x)(x2 + 9)1

2 d x

=12

⋅(x2 + 9)3

2

32

+ C

=13

(x2 + 9)32 + C

67

x3

= tan θ

⟹ sec θ =x2 + 9

3

x2 + 9 x

3θ

65. ∫ sec2 x tan3 x d x


Observe that dd x

(tan x) = sec2 x, hence we can immediately apply

the reverse chain rule.

∫ sec2 x tan3 x d x =tan4 x

4+ C

66. ∫ x2e−x d x


Our integrand is a product of two functions, so we integrate by parts. Letting:

∫ x2e−x d x = − x2e−x + 2∫ xe−x d x

Our new integral is again a product of two functions, so we integrate by parts again. Letting:

= − x2e−x + 2 [−xe−x + ∫ e−x d x] = − x2e−x − 2xe−x − 2e−x + C

= − e−x(x2 + 2x + 2) + C

68

u x2

e−x2x

−e−x

u′�

v

v′�

u x

e−x1

−e−x

u′�

v

v′�

67. ∫ xex2 d x


Observe that dd x

(x2) = 2x, so we can manipulate in order to apply


∫ xex2 d x =12 ∫ (2x)ex2 d x

=12

ex2 + C

68. ∫ sin x tan x d x

Method 1: Pythagorean Identity

We rewrite tan x =sin xcos x

and then apply the Pythagorean identity

sin2 θ + cos2 θ = 1.

∫ sin x tan x d x = ∫sin2 xcos x

d x

= ∫1 − cos2 x

cos xd x

= ∫ (sec x − cos x) d x

= ln |sec x + tan x | − sin x + C

69

69. ∫ sin4 x cos3 x d x

Method 1: Pythagorean Identity and Reverse Chain Rule

We observe that dd x

(sin x) = cos x, and that the power of cos x is

odd, so if we use the Pythagorean identity cos2 x = 1 − sin2 x, we will be in a good position to apply the reverse chain rule.

∫ sin4 x cos3 x d x = ∫ sin4 x(1 − sin2 x)cos x d x

= ∫ sin4 x cos x d x − ∫ sin6 x cos x d x

=15

sin5 x −17

sin7 x + C,

by the reverse chain rule.

70. ∫x3 + 1x3 − x

d x


We use an algebraic manipulation in order to simplify the integrand before we decompose into partial fractions.

∫x3 + 1x3 − x

d x = ∫x3 − x + x + 1

x3 − xd x

= ∫ (1 +x + 1

x(x2 − 1) ) d x

= ∫ (1 +1

x(x − 1) ) d x

Now, let 1x(x − 1)

=Ax

+B

x − 1.

∴ 1 = A(x − 1) + Bx

letting x = 1: B = 1

letting x = 0: A = − 1

Therefore 1x(x − 1)

= −1x

+1

x − 1.

Hence, ∫x3 + 1x3 − x

d x = ∫ (1 −1x

+1

x − 1 ) d x

= x − ln |x | + ln |x − 1 | + C

70

71. ∫ ln (x + x2 − 1) d x


Much like how we evaluate ∫ ln x d x, we will integrate by parts.

Letting:

∫ ln (x + x2 − 1) d x = x ln (x + x2 − 1) − ∫x

x2 − 1d x

We manipulate such that we can apply the reverse chain rule:

= x ln (x + x2 − 1) −12 ∫ (2x)(x2 − 1)− 1

2 d x

= x ln (x + x2 − 1) − x2 − 1 + C

72. ∫d x

(x + 1)12 + (x + 1)


There are no obvious substitutions, instead we replace the “most difficult” part of the integrand, which in this case is x + 1.

∫d x

(x + 1)12 + (x + 1)

= ∫2u

u + u2du

= 2∫du

u + 1

= 2 ln |u + 1 | + C


= 2 ln x + 1 + 1 + C

Section 8

Questions 71 - 80

71

u ln (x + x2 − 1)

11

x2 − 1

x

u′�

v

v′� let u = x + 1

dud x

=1

2 x + 1

d x = 2u du

73. ∫4

0

x

x + 4d x


Our goal is to rewrite the numerator in terms of x + 4.

∫4

0

x

x + 4d x = ∫

4

0

x + 4 − 4

x + 4d x

= ∫4

0 ( x + 4 −4

x + 4d x)

= [ 23

(x + 4)32 − 8 x + 4]

4

0

= ( 23

⋅ 832 − 8 8) − ( 2

3⋅ 4 3

2 − 8 4) =

32 23

− 16 2 −163

+ 16

=323

−16 2

3

=163 (2 − 2)


While we could substitute u = x + 4, u = x + 4 will be a much better substitution.

∫4

0

x

x + 4d x = ∫

2 2

2

u2 − 4u

(2u du)

= 2∫2 2

2(u2 − 4) du

= 2 [ 13

u3 − 4u]2 2

2

= 2 [( 163

2 − 8 2) − ( 83

− 8)] = 2 ( 16

3−

83

2) =

163 (2 − 2)

72

let u = x + 4

⟺ x = u2 − 4

dud x

=1

2 x + 4

d x = 2u du

when x = 0, u = 2

when x = 4, u = 2 2

74. ∫2

1

d xx(1 + x2)


Our denominator consists of a linear and quadratic factor, so we use partial fractions.

We decompose into partial fractions.

Let 1x(1 + x2)

=Ax

+Bx + Cx2 + 1

.

∴ 1 = A(x2 + 1) + (Bx + C )x

letting x = 0: A = 1

letting x = i: 1 = (Bi + C )i ⟹ − B + Ci = 1

Equating real and imaginary parts: B = − 1 and C = 0.

Therefore 1x(x2 + 1)

=1x

−x

x2 + 1

Hence ∫2

1

d xx(1 + x2)

= ∫2

1 ( 1x

−x

x2 + 1 ) d x

= [ln |x | −12

ln |x2 + 1 |]2

1

= (ln 2 −12

ln 5) − (0 −12

ln 2)

=32

ln 2 −12

ln 5

=12

ln85


A far more obscure substitution, but one that is very succinct is

the substitution u =1x

.

∫2

1

d xx(1 + x2)

= ∫12

1

u

1 + 1u2

(−1u2 ) du

= ∫1

12

u3

u2 + 1⋅

1u2

du

= ∫1

12

uu2 + 1

du

Manipulating such that we can apply the reverse chain rule:

=12 ∫

1

12

2uu2 + 1

du

=12 [ln |u2 + 1 |]

1

12

=12 (ln 2 − ln

54 )

=12

ln85

73

let u =1x

dud x

= −1x2

d x = −1u2

du

when x = 1, u = 1

when x = 2, u =12

75. ∫2

1

ln xx

d x


Observe dd x

(ln x) =1x

, thus we apply the reverse chain rule.

∫2

1

ln xx

d x = [ 12

(ln x)2]2

1

=12

(ln 2)2 − 0

=12

(ln 2)2

76. ∫1

0cos−1 x d x


This integral we deal with by integrating by parts, where v′� = 1. Letting:

∫1

0cos−1 x d x = [x cos−1 x]

1

0+ ∫

1

0

x

1 − x2d x

We then manipulate such that we can apply the reverse chain rule:

= [x cos−1 x]1

0−

12 ∫

1

0

(−2x)

1 − x2d x

= [x cos−1 x]1

0− [ 1 − x2]

1

0

= (0 − 0) − (0 − 1)

= 1

74

u cos−1 x

1−1

1 − x2

x

u′�

v

v′�

77. ∫2

1

x + 1

−2 + 3x − x2d x


Our goal is to manipulate the numerator and express it in terms of the derivative of the bottom.

As x + 1 = −12

(−2x + 3) +52

:

∫2

1

x + 1

−2 + 3x − x2d x = −

12 ∫

2

1

2x

−2 + 3x − x2+

52 ∫

d x

−2 + 3x − x2

We apply the reverse chain rule to the left most integral, and complete the square to the right most integral.

= − [ −2 + 3x − x2]2

1+

52 ∫

1

14 − (x − 3

2 )2

d x

= − (0 − 0) +52

sin−1x − 3

212

2

1

=52 [sin−1(2x − 3)]

2

1

=52 (sin−1 1 − sin−1(−1))

=5π2

78. ∫π2

0



Before we apply the t-substitution, we need to simplify. We do so using the Pythagorean identity and double angle formula.

∫π2

0


= lima→ π

2− ∫

a

0


d x

= lima→ π

2− ∫

a

0

d x

1 + 12 (1 − cos 2x)

d x

= 2 lima→ π

2− ∫

a

0

d x3 − cos 2x

Now we are able to apply the t-substitution.

= 2 lima→ π

2− ∫

tan a

0

11 + t2

3 − 1 − t2

1 + t2

dt

= 2 lima→ π

2− ∫

tan a

0

12 + 4t2

dt

= lima→ π

2− ∫

tan a

0

11 + 2t2

dt

= lima→ π

2− [ 1

2tan−1 ( 2t)]

tan a

0

75

let t = tan x

d x =1

1 + t2dt

when x = 0, t = 0

when x = a, t = tan a

=1

2lim

a→ π2

−tan−1 ( 2 tan a)

Now as a →π2

−, 2 tan a → + ∞, hence tan−1 ( 2 tan a) →

π2

−.

=1

2⋅

π2

=π 2

4


∫π2

0


= lima→ π

2− ∫

a

0


d x

We divide the top and bottom by cos2 x.

= lima→ π

2− ∫

a

0

sec2 x1 + 2 tan2 x

d x

= lima→ π

2−

1

2 ∫a

0

2 sec2 x

1 + ( 2 tan x)2 d x

= lima→ π

2−

1

2 [tan−1 ( 2 tan x)]a

0

=1

2lim

a→ π2

−tan−1 ( 2 tan a)

=π 2

4 (by the same argument above)

79. ∫1

0x 1 − x2 d x


Observe that dd x

(1 − x2) = − 2x, so we manipulate such that we


∫1

0x 1 − x2 d x = −

12 ∫

1

0(−2x) 1 − x2 d x

= −12

⋅23 [(1 − x2)3

2]1

0

= −13

(0 − 1)

=13

76

80. ∫4

2x ln x d x


Our integrand is a product of two integrals, so we apply integration by parts. Letting:

∫4

2x ln x d x = [ 1

2x2 ln x]

4

2−

12 ∫

4

2x d x

= ( 162

ln 4 −42

ln 2) −12 [ x2

2 ]4

2

= (16 ln 2 − 2 ln 2) −12

(8 − 2)

= 14 ln 2 − 3

u ln x

x1x

12

x2

u′�

v

v′�

77

81. ∫2

1

d xx2 + 5x + 4


Factorising the denominator into two linear factors, we then decompose the integrand into partial fractions.

∫2

1

d xx2 + 5x + 4

= ∫2

1

d x(x + 4)(x + 1)

Let 1(x + 4)(x + 1)

=A

x + 4+

Bx + 1

.

∴ A(x + 1) + B(x + 4) = 1

letting x = − 1: 3B = 1 ⟹ B =13

letting x = − 4: −3A = 1 ⟹ A = −13

Therefore 1(x + 4)(x + 1)

= −1

3(x + 4)+

13(x + 1)

.

Hence:

∫2

1

d xx2 + 5x + 4

= ∫2

1 (−1

3(x + 4)+

13(x + 1) ) d x

=13 [ln |x + 1 | − ln |x + 4 |]2

1

=13 [(ln 3 − ln 2) − (ln 6 − ln 5)]

=13

ln ( 3 × 52 × 6 )

=13

ln ( 54 )

Section 9

Questions 81 - 90

78

82. ∫π2

0 (1 +12

sin x)−1

d x


There are no immediate substitutions or manipulations, and our integrand contains a single trigonometric function, hence we apply the t-substitution.

∫π2

0(1 +

12

sin x)−1 d x = ∫π2

0

22 + sin x

d x

= ∫1

0

2

2 + 2t1 + t2

×2

1 + t2dt

= 2∫1

0

dt1 + t + t2

We then complete the square for the quadratic in the denominator:

= ∫1

0

234 + (t + 1

2 )2 dt

=4

3tan−1

t + 12

32

1

0

=4

3tan−1 ( 2t + 1

3 )1

0

=4

3 (tan−1 3 − tan−1 1

3 ) =

4

3 ( π3

−π6 )

=2π

3 3

79

83. ∫1

0x2e−x d x


Our integrand is a product of two functions, so we apply integration by parts. Letting:

∫1

0x2e−x d x = [−x2e−x]1

0+ 2∫

1

0xe−x d x

Our new integrand is again a product of two functions, so we integrate by parts again. Letting:

= [− x2e−x]1

0+ 2 [[ − xe−x]

1

0+ ∫

1

0e−x d x]

= [− x2e−x − 2xe−x − 2e−x]1

0

= − [e−x(x2 + 2x + 2)]1

0

= 2 − 5e−1

84. ∫1

0

7 + x1 + x + x2 + x3

d x


Factorising the denominator into a linear and quadratic factor, we then decompose the integrand into partial fractions.

∫1

0

7 + x1 + x + x2 + x3

d x = ∫1

0

7 + x(x + 1)(x2 + 1)

d x

Now, let 7 + x(x + 1)(x2 + 1)

=A

x + 1+

Bx + Cx2 + 1

.

∴ A(x2 + 1) + (Bx + C )(x + 1) = 7 + x

letting x = − 1: 2A = 6 ⟹ A = 3

letting x = 0: A + C = 7 ⟹ C = 4

letting x = 1: 2A + 2B + 2C = 8 ⟹ B = − 3

Therefore 7 + x(x + 1)(x2 + 1)

=3

x + 1+

−3x + 4x2 + 1

Hence:∫1

0

7 + x1 + x + x2 + x3

d x = ∫1

0 ( 3x + 1

−3x

x2 + 1+

4x2 + 1 ) d x

= [3 ln |x + 1 | −32

ln |x2 + 1 | + 4 tan−1 x]1

0

=32

ln 2 + π

80

u x2

e−x2x

−e−x

u′�

v

v′�

u x

e−x1

−e−x

u′�

v

v′�

85. ∫1

0

e−2x

e−x + 1d x


There are no immediate substitutions/manipulations, so a good substitution to try would be u = ex.

∫1

0

e−2x

e−x + 1d x = ∫

e

1

u−2

u−1 + 1⋅ u−1 du

= ∫e

1

u−3

u−1 + 1du

= ∫e

1

duu3 + u2

= ∫e

1

duu2(1 + u)

Our denominator consists of a linear and quadratic factor, so we decompose into partial fractions.

Let 1u2(1 + u)

=Au

+Bu2

+C

u + 1.

∴ 1 = Au(u + 1) + B(u + 1) + Cu2

letting u = 0: B = 1

letting u = − 1: C = 1

letting u = 1: 1 = 2A + 2B + C ⟹ A = − 1

Therefore: 1u2(1 + u)

= −1u

+1u2

+1

u + 1

Hence: ∫1

0

e−2x

e−x + 1d x = ∫

e

1 (−1u

+1u2

+1

u + 1 ) du

= [−ln u −1u

+ ln |1 + u |]e

1

= [ln1 + u

u−

1u ]

e

1

= (ln1 + e

e−

1e ) − (ln 2 − 1)

= ln ( 1 + e2e ) −

1e

+ 1

81

let u = ex

d xdu

= ex

d x =1u

du

when x = 0, u = 1

when x = 1, u = e

86. ∫a2

0

ya − y

dy


Our goal is to express the numerator in terms of the denominator.

As y = − (a − y) + a:

∫a2

0

ya − y

dy = − ∫a2

0

a − ya − y

dy + ∫a2

0

aa − y

dy

= − ∫a2

01 dy − [a ln |a − y |]

a2

0

= − [y]a2

0− [a ln |a − y |]

a2

0

= −a2

− a (lna2

− ln |a |)

= −a2

− a ln( a

2 )a

= −a2

− a ln ( 12 )

= −a2

+ a ln 2

=a2

(ln 4 − 1)

87. ∫a

0

(a − x)2

a2 + x2d x


Expanding the numerator, we can manipulate the integral such that we can easily apply the reverse chain rule.

∫a

0

(a − x)2

a2 + x2d x = ∫

a

0

a2 − 2a x + x2

a2 + x2d x

= ∫a

0 (1 −2a x

a2 + x2 ) d x

= [x − a ln |a2 + x2 |]a

0

= (a − a ln |2a2 |) − (0 − a ln |a2 | )

= a − a ln2a2

a2

= a(1 − ln 2)

82

88. ∫1

0

x + 3(x + 2)(x + 1)2

d x


Our denominator consists of a linear and quadratic factor, so we decompose the integrand into partial fractions.

Let x + 3(x + 2)(x + 1)2

=A

x + 2+

Bx + 1

+C

(x + 1)2.

∴ A(x + 1)2 + B(x + 1)(x + 2) + C(x + 2) = x + 3

letting x = − 1: C = 2

letting x = − 2: A = 1

letting x = 0: A + 2B + 2C = 3 ⟹ B = − 1

Therefore x + 3(x + 2)(x + 1)2

=1

x + 2−

1x + 1

+2

(x + 1)2

Hence: ∫1

0

x + 3(x + 2)(x + 1)2

d x = ∫1

0 ( 1x + 2

−1

x + 1+

2(x + 1)2 ) d x

= [ln |x + 2 | − ln |x + 1 | −2

x + 1 ]1

0

= (ln 3 − ln 2 − 1) − (ln 2 − 0 − 2)

= ln34

+ 1

89. ∫1

0

x2

x6 + 1d x


Observe that dd x

(x3) = 3x2, thus we can manipulate such that we


∫1

0

x2

x6 + 1d x =

13 ∫

1

0

3x2

(x3)2 + 1d x

=13 [tan−1(x3)]

1

0

=13 ( π

4− 0)

=π12

83

90. ∫π

0cos2 m x d x, where m is an integer.

Method 1: Half Angle Formula

We aim to simplify the integral by reducing the power of cos(m x).

We do so by applying the half angle formula cos2 x =12

(1 + cos 2x).

∫π

0cos2 m x d x =

12 ∫

π

0(1 + cos 2m x) d x

=12 [x −

sin 2m x2m ]

π

0

=12 (π −

12m

sin(2mπ)) − 0

Now the sine of any integer multiple of π is zero, hence:

∫π

0cos2 m x d x =

π2

84

91. ∫π2

π4

x sin 2x d x


Our integrand is a product of two functions, hence we integrate by parts. Letting:

∫π2

π4

x sin 2x d x = [−12

x cos 2x]π2

π4

+12 ∫

π2

π4

cos 2x d x

= [−12

x cos 2x +14

sin 2x]π2

π4

= ( π4

+ 0) − (0 −π4 )

=π4

−14

92. ∫a2

0x2 a2 − x2 d x


We can use the Pythagorean identity 1 − sin2 θ = cos2 θ to help simplify the integral.

∫a2

0x2 a2 − x2 d x = ∫

π6

0(a sin θ)2 a2 − a2 sin2 θ(a cos θ) dθ

= a4 ∫π6

0sin2 θ cos2 θ dθ

We then use the double and half angle formula to help simplify the integral.

Section 10

Questions 91 - 100

85

u x

sin 2x1

−12

cos 2x

u′�

v

v′�

let x = a sin θ when x = 0, θ = 0

d xdθ

= a cos θ when x = a, θ =π6

d x = a cos θ dθ

=a4

4 ∫π6

0(2 sin θ cos θ)2 dθ

=a4

4 ∫π6

0sin2 2θ dθ

=a4

8 ∫π6

0(1 − cos 4θ) dθ

=a4

8 [θ −14

sin 4θ]π6

0

=a4

8 ( π6

−14

sin2π3 ) − 0

=a4

8 ( π6

−3

8 ) =

a4

192 (4π − 3 3)

93. ∫π4

0sec2 x tan x d x


Observe that dd x

(tan x) = sec2 x, hence we can immediately apply


∫π4

0sec2 x tan x d x = [ 1

2tan2 x]

π4

0

= ( 12

− 0) =

12

86

94. ∫1

0(x + 2) x2 + 4x + 5 d x


Observe that dd x

(x2 + 4x) = 2x + 4, hence we manipulate such

that we can apply the reverse chain rule.

∫1

0(x + 2) x2 + 4x + 5 d x =

12 ∫

1

0(2x + 4) x2 + 4x + 5 d x

=12

⋅23 [(x2 + 4x + 5)3

2]1

0

=13 (10 3

2 − 532)

=13 (10 10 − 5 5)

=5 5

3 (2 2 − 1)

95. ∫2

1x(ln x)2 d x


Our integrand is a product of two functions, hence we integrate by parts. Letting:

∫2

1x(ln x)2 d x = [ x2

2(ln x)2]

2

1

− ∫2

1x ln x d x

Our new integrand is again a product of two functions, hence we integrate by parts again. Letting:

= [ x2

2(ln x)2]

2

1

− ([ 12

x2 ln x]2

1−

12 ∫

2

1x d x)

87

u (ln x)2

x2 ln x

x

x2

2

u′�

v

v′�

u ln x

x1x

x2

2

u′�

v

v′�

∫2

1x(ln x)2 d x = [ 1

2x2(ln x)2 −

12

x2 ln x −14

x2]2

1

= [ 12

⋅ 4(ln 2)2 −12

⋅ 4 ln 2 −14

⋅ 4] − (0 − 0 −14 )

= 2(ln 2)2 − 2 ln 2 +34

96. ∫4

3

x2 + 4x2 − 1

d x


We simplify by rewriting the numerator in terms of the denominator, and then decomposing into partial fractions.

∫4

3

x2 + 4x2 − 1

d x = ∫4

3 ( x2 − 1x2 − 1

+5

x2 − 1 ) d x

Now, let 1x2 − 1

=A

x + 1+

Bx − 1

.

∴ A(x − 1) + B(x + 1) = 1

letting x = 1: 2B = 1 ⟹ B =12

letting x = − 1: −2A = 1 ⟹ A = −12

Therefore 1x2 − 1

= −1

2(x + 1)+

12(x − 1)

Hence: ∫4

3

x2 + 4x2 − 1

d x = ∫4

3 (1 −1

2(x + 1)+

12(x − 1) ) d x

= [x −12

ln |x + 1 | +12

ln |x − 1 |]4

3

= (4 −12

ln 5 +12

ln 3) − (3 −12

ln 4 +12

ln 2) = 1 +

52

ln65

88

97. ∫4

1

x2 + 4x(x + 2)

d x


We first simplify by expressing the numerator in terms of the denominator x2 + 2x:

∫4

1

x2 + 4x(x + 2)

d x = ∫4

1

x2 + 2x − 2x + 4x(x + 2)

d x

= ∫4

1 (1 −2x − 4

x(x + 2) ) d x

We then decompose the integrand into partial fractions.

Let 2x − 4x(x + 2)

=Ax

+B

x + 2

∴ A(x + 2) + Bx = 2x − 4

letting x = 0: 2A = − 4 ⟹ A = − 2

letting x = − 2: −2B = − 8 ⟹ B = 4

Therefore 2x − 4x(x + 2)

= −2x

+4

x + 2

Hence:

∫4

1

x2 + 4x(x + 2)

d x = ∫4

1 (1 +2x

−4

x + 2 ) d x

= [x + 2 ln |x | − 4 ln |x + 2 |]4

1

= (4 + 2 ln 4 − 4 ln 6) − (1 + 0 − 4 ln 3)

= 3 + ln ( 42 × 34

64 ) = 3 + ln 1

= 3

89

98. ∫π2

0

cos x5 − 3 sin x

d x


Observe that dd x

(5 − 3 sin x) = − 3 cos x, hence we manipulate

such that we can apply the reverse chain rule.

∫π2

0

cos x5 − 3 sin x

d x = −13 ∫

π2

0

−3 cos x5 − 3 sin x

d x

= [−13

ln |5 − 3 sin x |]π2

0

= −13 (ln 2 − ln 5)

=13

ln52

99. ∫1

0

1

(4 − x2)32

d x


We can make use of the Pythagorean identity 1 − sin2 θ = cos2 θ to help simplify the integrand:

∫1

0

1

(4 − x2)32

d x = ∫π6

0

2 cos θ

(4 − 4 sin2 θ)32

dθ

= ∫π6

0

2 cos θ8 cos3 θ

dθ

=14 ∫

π6

0sec2 θ dθ

=14 [tan θ]

π6

0

=14 ( 1

3− 0)

=3

12

90

let x = 2 sin θ

d xdθ

= 2 cos θ

d x = 2 cos θ dθ

when x = 0, θ = 0

when x = 1, θ =π6

100.∫π2

02 sin θ cos θ(3 sin θ − 4 sin3 θ) dθ

Method 1: Double/Triple Angle Identity and Product to Sums

We simplify the integrand by recognising that:

sin(2θ) = 2 sin θ cos θ and sin(3θ) = 3 sin θ − 4 sin3 θ

∫π2

02 sin θ cos θ(3 sin θ − 4 sin3 θ) dθ = ∫

π2

0sin 2θ sin 3θ dθ

Our integrand is now a product of two trigonometric functions, so we apply the product to sum identities.

cos(A − B) − cos(A + B) = 2 sin A sin B

⟹ sin A sin B =12 [cos(A − B) − cos(A + B)]

Letting A = 3θ and B = 2θ in our identity above:

∫π2

02 sin θ cos θ(3 sin θ − 4 sin3 θ) dθ =

12 ∫

π2

0(cos θ − cos 5θ) dθ

=12 [sin θ −

15

sin 5θ]π2

0

=12 (1 −

15 )

=25

91

Documents

Coroneos' 100 Integrals - · PDF fileChapter 1 The 100 Integrals The ideal way to use this book is to: • Do a section of ten in one go. (do NOT look at the solutions) • Mark your