4.UNIT 4 INTEGRATION

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UNIT 4: INTEGRATION
______________________________________________

INTRODUCTION

Every operation in mathematics has its inverse. In this chapter we learn how to
reverse the process of differentiation with the process of integration.

LEARNING OUTCOME

After completing this unit, students should be able to:
1. Find the integrals of functions by considering integration as a reverse of
differentiation.
2. Integrate indefinite integrals by rule of integration.
3. Integrate composite functions by using substitution method.
4. Integrate product of functions using integration by part.
5. Integrate quotient of functions using partial fraction method.
6. Evaluate the definite integrals.

4.1 INTEGRATION AS THE REVERSE OF DIFFERENTIATION

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Integration is the reverse process of differentiation. The process of obtaining

dy

from y (a function of a) is known as differentiation. Hence, the process of

dx
dy

obtaining y from is known as integration. Integration of y with respect to x, is

dx

 denoted by f(x) dx . The symbols f(x) dx denote the integral of f(x) with

respect to the variable x. For example:

y = x3 + c Differentiation dy = x2
3 Integration dx

4.2 INDEFINITE INTEGRAL OF FUNCTION

From which the derivative 5x4 was derived?. Take a look at these examples:

( )d x5 = 5x4  5x4 dx = x5
 5x4 dx = x5 + 3
dx  5x4 dx = x5 − 1

( )also d x5 + 3 = 5x4
dx

( )and d x5 −1 = 5x4
dx

Any constant terms in the original expression becomes zero in the derivative.
Therefore the presence of such constant term is replaced by adding a symbol c to
the result of integration.

 f'(x) dx = f(x) + c

c is known as the constant of integration and must always be included. Such integral
is called an indefinite integral.

4.2.1 INTEGRATION OF POLYNOMIALS

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1.  a dx = ax + c where a is a constant 4. (ax + b)n dx = (ax + b)n + 1 +c

2. xn dx = xn + 1 + c (n + 1) d (ax + b)
dx
(n + 1) (n  −1)
; (n  −1)

3. axn dx = a xn + 1 + c (n  −1)   5. f(x)  g(x) dx = f(x) dx  g(x) dx

(n + 1)

Example 4.2.1.1 : Integrating a constant

Integrate the following with respect to x :

(a) 2 (b) −5 (c) 1
3

Solution :

(a) 2 dx = 2x + c

(b) − 5 dx = −5x + c
(c) 1 dx = 1 x + c

33

Example 4.2.1.2 : Integration of xn
Integrate the following with respect to x :

(a) x 5 (b) 1 (c) 1
x3 x

Solution :  x 5 dx = 1 x 5+1 + c = 1 x 6 + c
(a) 5 +1 6
(b)
 1 dx = x −3 dx = − 1 x −2 + c = − 1 + c
(c) x 3 2 2x 2

 1 −1 1 − 1 +1 1

dx = x 2 dx = x 2 + c = 2x 2 + c = 2 x + c
x − 1 +1
2

Example 4.2.1.3 : Integration of axn

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Find the following integrals:

(a) -4x2 (b) 2x3 (c) 3x5

Solution :  − 4x 2 dx = −4 x2 dx = −4  x3  + c = − 4 x 3 +c
(a) 3 3
(b)
(c)  2x 3 dx = 2 x3 dx = 2  x4  + c = 1 x4 +c
4 2

 3x 5 dx = 3 x5 dx = 3  x6  + c = 1 x 6 +c
6 2

Example 4.2.1.4 : Integral of (ax + b)n
Find:

(a)  (2x + 3)4 dx (b)  (3 − 6x)−4 dx (c)  (1− x)− 1 dx
2

Solution :

(a)  (2x + 3)4 dx = 1 (2x + 3)5 + c = 1 (2x + 3)5 + c
10
(5) (2)

(b)  (3 − )6x −4 dx = (− 1 6) (3 − )6x −3 + c = 1 (3 − )6x −3 + c
18
3) (−

(c)  (1 − x )− 1 dx = 1 (1− )1 + c = −2 (1− )1 + c
2
 1  (− 1) x2 x2

2

Example 4.2.1.5 : Integral of sum and differences

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Find the following integrals:

(a)   x2 + 3 + 1  dx (b)  (2x −1)2 dx (c)  x+ 1 dx
 x3  x3

Solution :     x 2 + 3 + 1  dx = x 2 dx + 3 1 dx + x −3 dx
(a)  x3 
= 1 x 3 + 3x − 1 x −2 + c
32

  ( )(b) (2x − 1)2 dx = 4x 2 − 4x + 1 dx = 4 x 3 − 2x 2 + x + c
3

  ( )(c) x + 1 dx = x −2 + x −3 dx = x −1 + x −2 + c = − 1 − 1 + c
x3 −1 −2 x 2x 2

EXERCISE 4.2.1 b) 6 dx
x4
1. Find the following integrals:

a) −15 dx

( )c) 3x2−4x3 dx d) 3 − 4 + 6  dx
 x2 x3 

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e) (1 + 4x − 6x2 ) dx f)  x + 1  dx
 x2 

g) (3x + 2)(2x −1)dx h) 3x3 − 3x2 + 4 dx
x2

Answers: a) -15x + c, b) − 2 +c , c) x3 − x4 + c , d) 3x + 4 − 3 + c ,
x3 x x2

e) x + 2x2 − 2x3 + c , f) 1 x2 − 1 +c , g) 2x3 + x2 − 2x + c , h) 3 x2 −3x − 4 +c
2 2x
2x

2. Find the following integrals: b)  (x − 9)8dx

a)  (x + 1)3 dx

c) (4x + 7)5 dx d) (3x − 8)6 dx

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e) 2(4 − 3x) −6 dx f) 5 dx
(4 − 3x)3

Answers: a) 1 (x + 1)4 + c , b) 1 (x − 9)9 +c , c) 1 (4x + 7)6 +c ,
d) 4
9 24
1 (3x −8)7 + c , e)
2 + c , f) 6(4 5 +c
21 15(4 − 3x)5 − 3x)2

1
4.2.2 INTEGRATION OF x

➢ 1 dx = ln x + c d ln u = 1 • d (u)
x
dx u dx

➢ 1 dx = ln ax + b +c

ax + b d (ax + b)

dx

Example 4.2.2 : (b) 1 dx (c) 3 dx
Find 2x + 3 5 − 2x

(a) 5 dx
x

Solution :  5 dx = 5 ln x + c
(a) x
(b)
(c)  1 dx = 1 ln 2x + 3 + c
2x + 3 2

 3 dx = − 3 ln 5 − 2x + c
5 − 2x 2

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EXERCISE 4.2.2 b) 1 dx
2x
Find the following integrals:

a)  3 dx
x

c) − 5 dx d) 7 dx
x 2x

3 f) 1 dx
2x + 1
e) 2 dx
3x

g) 4 dx h) − 10 dx
3x + 2 6 − 5x

Answers: a) 3 ln x + c, b) 1 ln x + c , c) – 5 ln x + c , d) 7 ln x + c , e) 1 ln x + c ,
2 22

f) 1 ln (2x + 1) + c , g) 4 ln (3x + 2) + c , h) 2 ln (6 − 5x) + c
23

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4.2.3 INTEGRATION OF TRIGONOMETRIC FUNCTIONS

 sin(ax n ) dx = − cos(ax n ) +c
d (ax n )
dx
d sin(axn ) = cos(axn ) d (axn )
 cos(ax n ) dx = sin(ax n ) +c dx dx
d (ax n ) d cos(axn ) = − sin(axn ) d (axn )
dx dx dx
d tan (axn ) = sec2 (axn ) d (axn )
dx dx

 sec2 (ax n ) dx = tan (ax n ) + c
d (ax n )
dx

Example 4.2.3 :

Find the following integrals :

(a) sin 3x (b) 3cos 2x (c) 2cos (3x − 1)

(d) sec 2 (1− 4x) dx (e) cos 3x − sin 1 x
2

Solution :

(a)  sin 3x dx = − 1 cos 3x + c
3

(b)  3 cos 2x dx = 3 cos 2x dx = 3  1  sin 2x + c = 3 sin 2x + c
 2 2

(c)  2cos (3x − 1) dx = 2 cos (3x − 1) dx = 2 sin(3x −1) + c
3

(d)  sec2 (1− 4x) dx = − 1 tan (1− 4x) + c
4

(e) cos 3x − sin 1 x dx = 1 sin 3x −  1 cos 1 x  +c = 1 sin 3x + 2 cos 1x+c
2 3  −1  3 2
 2 
2

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EXERCISE 4.2.3 b) cos 4x dx
Find the following integrals:

a) sin 2x dx

c) sec 2 5x dx d) sin 1 x dx
e) sin 2 x dx 2

3 f) sec 2 3 x dx
4

g)  sin (3x + ) dx h) cos 5x −  dx
 2

i) sec 23x +  dx j) sin  3 x −  dx
 6 4 2

Answers: a) − 1 cos2x +c , b) 1 sin 4x + c , c) 1 tan 5x + c , d) − 2 cos 1 x + c ,
2 4 5 2

e) − 3 cos 2 x + c , f) 4 tan 3 x + c , g) − 1 cos( 3x + ) + c ,
23 34 3

h) 1 sin( 5x −  ) + c , i) 1 tan( 3x +  ) + c ,j) − 4 cos( 3 x −  ) + c
5 2 3 6 3 42

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4.2.4 INTEGRATION OF EXPONENTIAL FUNCTION

➢ ex dx = ex + c

➢ eax + b dx = eax + b +c d (ex ) = ex
dx
d (ax + b)
dx

Example 4.2.4 :

Find the int egral of the following functions :

(a) e −3x (b) 2e1−2x (c) ( )e−x + e x 2
(d) 5e x (e) 1

e2x

Solution :

(a) e−3x dx = − 1 e−3x + c
3

(b) 2e1−2x dx = 2 − 1 e1−2x  + c = − e1−2x + c
2 

(c) (e−2x + 2 + e2x ) dx = − 1 e−2x + 2x + 1 e2x + c
22

 (d) 5 ex dx = 5 ex dx =5 ex + c

 (e) 1 e−2x dx = − 1 e−2x + c
e2x dx = 2

EXERCISE 4.2.4

Find the integral with respect to x of:

a) e4x dx b)  e−5x dx

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1x −1x

c) e3 dx d) e 5 dx

e) e3x+2 dx f) e−4x+5 dx

3−4x ( )( )h) e2x + 3 e−x − 4 dx

g) e5 7 dx

Answers: a) 1 e4x + c , b) − 1 e−5x + c , c) 1x −1x
45
3 e3 + c , d) − 5e 5 + c

e) 1 e3x+2 + c , f) − 1 e−4x+5 +c, g) − 7 (3−4x) +c ,

34 4 e5 7

h) ex − 2e2x − 3e−x −12x + c

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4.3 TECHNIQUES OF INTEGRATION

To integrate functions which are not in standard form, we need to use a few
other techniques to make the process of integration easier. There are three
techniques:

4.3.1 Integration by Substitution
4.3.2 Integration by Part
4.3.3 Integration by Partial Fraction

4.3.1 INTEGRATION BY SUBSTITUTION Object
Image
 f(x) dx =  fg(u) g' (u) du

The substitution technique is simplified as follows:

a. Choose u = g(x)

b. Find du = g'(x)
dx

c. Substitute u = g(x), du = g ‘(x) dx. In this part, the integrand
must be in terms of u only, meaning there is no x terms left.
But if this happen, make another substitution for u.

d. Solve the integration
e. Back substitute u with g(x) therefore the final answer is in

terms of x.

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Example 4.3.1 :
By using a suitable substitution, find the following integration:

(a) (5x − 4)6 dx ( )(b) 3x 2 x3 − 1 dx

 dx 2 1

(c) (1 − x)3 (d) 9 x (1 − x3 )2 dx

Solution :

(a) (5x − 4)6 dx

Let u = 5x − 4  du = 5 and dx = 1 du
dx 5

  (5x − 4)6 dx = u6  1 du
5

= 1 u7  + c
5  
 7 

= (5x − 4)7 + c

35

( )(b) 3x 2 x3 − 1 dx

Let u = (x3 − 1)  du = 3x 2 dx

 3x 2 (x3 − 1)3 dx = u du

= u2 + c
2

= (x3 − 1)2 + c
2

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 dx

(c) (1 − x)3

Let u = (1 − x)  du = − dx

−du = dx

 dx = (1 − x) −3 dx
(1 − x) 3

= u −3 (− du)

= −  u −2  + c
 
 − 2 

= (1 − x) −2 + c
2

= 1 +c
2(1 − x) 2

2 1

(d) 9 x (1 − x3 )2 dx

Let u = 1 − x3  du = − 3x 2 dx

− 3du = 9x 2 dx

1

 9x 2(1 − x3 ) dx = u 2 (−3du)

1

= − 3 u 2 du

 3
 
= − 3 u 2  + c
3 
 2



3

= − 2(1 − x3 )2 + c

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EXERCISE 4.3.1

By using a suitable substitution, find the following integrals.

a) x2 (1 + x3 )4 dx b) x dx
1− 4x2

c) x(1 − x2 )1 dx ( )d) 4x 2x2 − 3 6 dx
2

e) dx 4x + 6
3−x x2 + 3x + 7 4 dx
 ( )f)

Answers: a) (1+ x3 )5 + c , b) − 3
15
1− 4x 2 + c , c) − (1− x2 )2 + c , d) (2x2 − 3)7 + c ,
4 37

e) − ln(3 − x) + c , f) −2 +c
3(x2 + 3x+ 7)3

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4.3.2 INTEGRATION BY PART

This is a method of integrating the product of two functions. The
product were either function not the derivative of the other, for

  example: x2 ln x dx, xex dx , ex sin x dx etc.

Consider the product rule for differentiation:

d (uv) = u dv dx + v du, u and v are functions of x.

dx dx dx

Integrate both sides with respect to x, WHEN & HOW TO USE

uv = u dv dx + v du dx, * If substitution doesn’t work.
dx dx *  f(x).g(x) dx match with

Rearranging the term,  u dv .
* choose u & dv and
u dv dx = uv − v du dx,
dx dx find du & v

 This is known as integration by part udv = uv − v du

The procedures:
a. Split the function into two simpler functions, one called u and

dv

the other .

dx

b. To decide which one will be u and dv.
(i) u should be a function which becomes a simpler function
after differentiation.
There is a simple acronym to remember which function
to equate u: L – P – E – T
1. L = Logarithm
2. P = Polynomial
3. E = Exponential
4. T = Trigonometry

dv

(ii) must be a function that is possible to integrate to obtain

dx

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Example 4.3.2 :

Find the following integrals using the integration by part techniques.

 (a) x cos x dx (b) x ln x dx (c) x2 ex dx

Solution :

(a) Let u = x dv = cos x dx

du = 1, v = cos x dx = sin x omit the integratio n constant
dx

 Substitutin g into udv = uv − v du gives

 x cos x dx = x sin x − sin x (1) dx

= x sin x − (− cos x) + c

= x sin x + cos x + c

(b) Let u = ln x dv = x dx

du = 1 , v = x dx = x2 omit the integratio n constant
dx x 2

 Substituti ng into udv = uv − v du gives

 x ln x dx = x2 ln x − x2  1 dx
2 2x

= x2 ln x − 1  1  x2 + c
2 22

= x2 ln x − x2 + c
24

(c) Let u = x2 dv = ex dx

du = 2x v = ex dx = ex omit the integratio n constant
dx

 Substituti ng into u dv = uv − v du gives

 x2 ex dx = x2 ex − ex  (2x) dx

= x2 ex − 2x ex dx again by part

u = 2x and dv = ex

du = 2 and v = ex dx = ex
dx
 = x2 ex − 2xe x − ex (2) dx

= x2 ex − −2xe x + 2ex + c

( )= ex x2 − 2x + 2 + c

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EXERCISE 4.3.2

Use the formula for differentiation by parts to complete the integration.

a) xex dx b) x cos 3x dx

c) xe −2x dx d) x sin x dx
2

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e) ln x dx f) x2ln x dx

2 h) ln 3x dx

g) x sin x dx

Answers: a) xe x − ex + c , b) x sin 3x + cos 3x + c , c) xe−2x − e−2x + c ,
39 −2 4

d) − 2x cos x + 4 sin x + c , e) x ln x − x + c , f) x3 ln x − x3 + c ,
22 39

g) –x2cos x + 2x sin x + 2 cos x + c, h) x ln 3x –x + c

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4.3.3 INTEGRATION BY PARTIAL FRACTION

In general, the integration in the form of f(x) dx where f(x) and g(x)
g(x)

are polynomials in terms of x, we have to express f (x ) as a partial
g(x)

fraction before integration is attempted. Only proper fractions can be
converted directly into partial fraction.

➢ nonrepeated linear factor (ax + b)  f(x) dx
g(x)
f(x) = A + B + C
(x + a)(x + b)(x + c) (x + a) (x + b) (x + c) STEPS
* determine the shape of the

partial fractions
( numbers of factor g ( x) )

* find value A, B & C
* evaluate the integral

Example 4.3.3 : dx (b)  (2x x+7 + 2) dx
Find:
− 3)(x
(a) 1
(x − 2)(x − 3)

Solution :

(a). Let 1 = A+B
(x − 2)(x − 3) x − 2 x − 3

So 1 = A(x − 3) + B(x − 2)

Substitute x = 2, 1 = − A  A = −1

Substitute x = 3 , 1 = B  B = 1

Thus 1 = −1 − 1
(x − 2)(x − 3) x − 2 x − 3

Hence  1 =   −1 + 1  dx
2)(x  x−2 − 
(x − − 3)  x 3 

= − ln x − 2 + ln x − 3) + c

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(b) Let (2x x+7 + 2) dx  A 3 + x B 2
2x − +
− 3)(x

x + 7  A (x + 2) + B (2x − 3)

Substitute x = −2 5 = −7B  B=−5
7

x=3 17 = 7 A  A = 17
2 22 7

Thus

17 5

(2x x+7 + 2)  7 3 − x 7 2
2x − +
− 3)(x

Hence

17 5

x+7 dx =  ( 7 3 − 7) dx
2x − x+2
 (2x − 3)(x + 2)

17 ln 2x −3 5 ln (x + 2)
7
= − +c
27

= 17 ln 2x − 3 − 5 ln (x + 2) + c

14 7

EXERCISE 4.3.3

Express the functions in each of the following integrals in partial fractions and hence

perform the integration.

a)  (1− 1 − 2) dx

x)(3x

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b)  (x 2x + 3 2) dx

− 4)(5x +

c) 3x dx
x2 − x − 2

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d) 3 dx
(x − 1)(2x − 1)

e) 1 dx
(4x − 1)(4x + 1)

Answers: a) -ln (1-x) + ln(3x-2) + c, b) 1 ln (x − 4) − ln(5x + 2) + c ,
2 10

c) 2 ln (x − 2) + ln (x + 1) + c , d) 3 ln (x − 1) − 3 ln( 2x − 1) + c , e) ln( 4x − 1) − ln( 4x + 1) + c
88

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4.4 THE DEFINITE INTEGRAL

An integral with limits is called a definite integral

b F x)
 f(x) F(b)− F(a)
dx = ( b =
a

a

b is known as the upper limit and a is known as the lower limit of the integral.

In the definite integral there is no constant of integration.

Example 4.4.1 :

Evaluate.

2 (b) ( )2 25

(a) 6x2 dx  1+ 2x + x2 dx (c) x dx
0 1
−1
(d) 3  x2 − 1  dx (e)
2 x2  −1 x − 1 2 dx

−2 x 

Solution :    ( )2  6x3 2 x3 2 = 2(8) = 16
(a) 6x 2  0
(b) 0
0
(c) dx =  3 = 2 = 2 23 − 0


 ( )2 dx =  + x2 + x3 2
1+ 2x + x2 x 3 
  −1
−1

= 2 + 4 + 8  −  − 1+ 1− 1 
 3  3

=9

25 1 3 25

dx = x 2 x2
 25x dx = 2

11 3 1

= 2 2532 − 1

3 

( )= 2 53 −1
3

= 82 2
3

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3  x2 − 1  dx = 3 x2 − x−2  x3 x−1 3  x3 13
  ( )(d) dx =  3 −  =  3 + 
2  x2   − 1   x 
2 2 2

= 1 (27) + 1 − 8 + 1
 3 3   3 2 

=61
6

−1 x − 1 2 −1  x3 x−1 −1

x2
  ( )(e)
−2 x  dx = − 2 + x−2 dx =  3 − 2x + − 1 
 
−2 − 2

=  x3 − 2x − 1  −1
 
 3 x  − 2

=  − 1 + 2 + 1 −  − 8 + 4 + 1 
3  3 2

= 2 2 −15
36

=5
6

EXERCISE 4.4

1. Evaluate the following definite integral.

9

a)  x dx
4

137

DUM20132 Date : ………………….

8

b) 3 x dx
1

c) 9 1 dx
1x

27 1 − 1
3
d) x dx
2
1

( )4

e) 6x − 3 x dx
1

Answers: a) 38 , b) 45 , c) 4, d) 6, e)31
34

138

DUM20132 Date : ………………….

2. Evaluate the following definite integral.

0

a)  (x + 1)(x + 2) dx
−2

b)3 x2 −1 dx
1 x2

c)2 3x 2 + 2 x dx
1 x2

139

DUM20132 Date : ………………….

2

d)  x (x − 1)(x − 2) dx
1

−2

e)  (3x − 1)(2x + 1) dx
−1

Answers: a) 2 4 1
, b) , c) 4.17, d) - , e)

3 3 4

140

DUM20132 Date : ………………….

SUMMARY

4.0
INTEGRATION

Definition. Integral of The integral of The definite Techniques of
polynomial trigonometric integral integration
Inverse of and exponent
differentiation and 1/x

 f'(x) dx = f(x) + c

Integration of xn.  cos x dx = sin x + c b By
 sin x dx = − cos x + c
 xn dx = 1 x n+1 +c  sec2 x dx = tan x + c  f (x) dx = F (x)ba Substitution
+
(n 1) a  f(x) dx =  fg(u) g'(u) du

= F (b)− F (a)

Integration of  cos (ax + b) dx = 1 sin (ax + b) + c By
a Part
 kx n dx udv = uv −  vdu
 sin (ax + b) dx = − 1 cos(ax + b)+ c
= kx n+1 + c a L-P-E-T
n +1
 sec2 (ax + b) dx = 1 tan (ax + b)+ c By
a Partial
Fraction
Integrating a ex dx = ex + c
constant  eax+b dx = 1 eax+b + c A+B +C
(x + a) (x + b) (x + c)
 a dx = ax + c a

 f '(x) e f (x) dx = e f (x) + c

Integral of
(ax + b)n

= (n 1 (ax + )b n+1 + c

+1)(a)

Integration of

1

x

= ln x + c

141