Calculus prepared b Engr.

James

Eastern Polytechnic, Port Harcourt

Department of Electrical/Electronics Engineering
MTH 122: Calculus

Lecturer: Engr. James Aduma John

Description: This course is intended to give the students a thorough knowledge of

limit of function differentiation and integration. This course covers topics such
differentiation and integration and their applications.

Prerequisites: Students should be familiar with the concept of differentiation and

integration of a function.

Learning outcomes: At the completion of this course, students are expected to:
i. understand the concept of derivative,
ii. differentiate explicit and implicit function,
iii. understand the concept of line, double and triple integrals,
iv. understand the concept of total derivative.
Assignments: We expect to have assignments, tests and a final examination for
this course. The assignments will be given at the end of a particular topic and a
tests will be administered and structured as preparation for the final examination.
Grading: We will assign 10% of this class grade to assignments, 20% for the test
and 70% for the final examination. The Final examination is comprehensive.

Textbook: The recommended textbook for this class are as stated:

Title: Engineering Mathematics, 7th Edition

Authors: K. A. Stroud and Dexter J. Booth.

Title: Higher Engineering Mathematics.

Authors: Dass, H. K. and Verma, E. R.

Title: Schaum’s Outline: Advanced Calculus, Fourth Edition.

Authors: Robert Wrede and Murray R. Spiegel.

Title; Calculus

Autors: P.U. Uzoma (Ph.D)

Calculus prepared b Engr. James

Real Valued Function

Peter Dirichlet a German Mathematician (1829) conceived a function as a variable,

called the dependent variable having its value fixed or determine in some definite
manner by the value assigned to the independent variable x or to several
independent variables x1, x2, … , x𝑛. The value of both 𝑦 and x are real. The
statement 𝑦 = 𝑓(x) is read as 𝑦 is a function of x. Again, 𝑦 = 𝑓(x1, x2, … , x𝑛)
indicates the interdependence between the variable 𝑦 and x. The function 𝑓(x) is
usually given as an explicit formula such as 𝑓(x) = x2 − 3x + 5 for all x real.

Usually, in algebraic expression, a real variable x may take any value in a certain
range. If the lowest value of x is a and the highest value of x is b and x may take
any value between a and b , then x is said to be a continuous variable in the range
[a , b] and takes all values such that a ≤ x ≤ b . Since the end points are included
among the values of x which form this range, the interval is called a closed
interval. The interval defined by the inequality a < x < 𝑏 is called an open interval
and is denoted by (a , b).

Review of Differentiation
Calculus prepared b Engr. James
Calculus prepared b Engr. James
Calculus prepared b Engr. James
Calculus prepared b Engr. James
Calculus prepared b Engr. James
Calculus prepared b Engr. James

1. Function of a Function.

If 𝑦 is a function of 𝑢 and that 𝑢 itself is a function of x, then the derivative of 𝑦

with respect to x is
𝑑y 𝑑y 𝑑𝑢
= ×
𝑑x 𝑑𝑢 𝑑x

This is also called the chain rule of differentiation.

Example1: Find the derivative of each of the following.

(a) 𝑦 = (3x2 − 2)3 (𝑏) 𝑦 = √(1 − 2x3)

Calculus prepared b Engr. James

5 1
(𝑐) 𝑦 = (𝑑) 𝑦 =
(6 − x2)3 √(1 + x2)
Solution

(a) 𝑦 = (3x2 − 2)3

Let = 3x2 − 2 , 𝑦 = 𝑢3
𝑑𝑢 𝑑𝑦
∴ = 6x, = 3𝑢2
𝑑x 𝑑𝑢
𝑑y 𝑑y 𝑑𝑢
= × = 3𝑢2 × 6x = 18x𝑢2 = 18x(3x2 − 2)2
𝑑x 𝑑𝑢 𝑑x

(b) 𝑦 = √(1 − 2x3)

Let 𝑢 = 1 − 2x3 , 𝑦 = 𝑢1⁄2

𝑑𝑢 𝑑𝑦 1 1
∴ = −6x2 , −1⁄2 =
= 𝑢
𝑑x 𝑑𝑢 2 2𝑢1⁄2
𝑑y 𝑑y 𝑑𝑢 1 −6x2 −3x2 −3x2
= × = × −6x =
2 = =
𝑑x 𝑑𝑢 𝑑x 2𝑢1⁄2 2𝑢1⁄2 √𝑢 √(1 − 2x3)
5
(c) 𝑦 =
(6 − x2)3
5
Let 𝑢 = 6 − x2 , 𝑦= = 5𝑢−3
𝑢3
𝑑𝑢 𝑑𝑦 −15
∴ = −2x , = −15𝑢−4 =
𝑑x 𝑑𝑢 𝑢4
𝑑y 𝑑y 𝑑𝑢 −15 30x 30x
= × = × −2x = 𝑢4 =
𝑑x 𝑑𝑢 𝑑x 𝑢4 (6 − x2)4
1
(d) 𝑦 =
√(1 + x2)
1
Let 𝑢 = 1 + x2 , 𝑦= = 𝑢−1⁄2
𝑢1⁄2
Calculus prepared b Engr. James
1 −1
∴ 𝑑𝑢 = 2x , 𝑑𝑦 −3/2 =
=− 𝑢
𝑑x 𝑑𝑢 2 2𝑢3⁄2
𝑑y 𝑑y 𝑑𝑢 −1 −x −x −x
= × = × 2x = = =
𝑑x 𝑑𝑢 𝑑x 2𝑢3⁄2 𝑢3⁄2 (1 + x2)3⁄2 √(1 + x2)3
2. Derivative of a Product.

If 𝑦 = 𝑢𝑣 where 𝑢 and 𝑣 are functions of x, then the derivative of 𝑦 with respect

to x is
𝑑y 𝑑v 𝑑𝑢
=𝑢 +𝑣
𝑑x 𝑑x 𝑑x
Example2: Find the derivative of each of the following.

(a) 𝑦 = (3 + 2x)(1 − x) (𝑏) 𝑦 = (1 − 2x + 3x2)(4 − 5x2)

(𝑐) 𝑦 = √x (1 + 2x)2 (𝑑) 𝑦 = (5x4 + 1)7𝑠𝑖𝑛3x

Solution

(a) 𝑦 = (3 + 2x)(1 − x)

Let = 3 + 2x , 𝑣 = 1 − x

𝑑𝑢 𝑑𝑣
= 2, = −1
𝑑x 𝑑x
𝑑y 𝑑v 𝑑𝑢
∴ =𝑢 +𝑣 = (3 + 2x)(−1) + (1 − x)(2) = −3 − 2x + 2 − 2x
𝑑x 𝑑x 𝑑x
= −1 − 4x

(b) 𝑦 = (1 − 2x + 3x2)(4 − 5x2)

Let = 1 − 2x + 3x2 , 𝑣 = 4 − 5x2

𝑑𝑢 𝑑𝑣
= −2 + 6x , = −10x
𝑑x 𝑑x
Calculus prepared b Engr. James

𝑑y 𝑑v 𝑑𝑢
∴ =𝑢 +𝑣 = (1 − 2x + 3x2)(−10x) + (4 − 5x2)(−2 + 6x)
𝑑x 𝑑x 𝑑x
= −10x(1 − 2x + 3x2) + (6x − 2)(4 − 5x2)

(c) 𝑦 = √x (1 + 2x)2

Let = √x = x1/2, 𝑣 = (1 + 2x)2

𝑑𝑢 1 𝑑𝑣
= , = 4(1 + 2x)
𝑑x 2x1/2 𝑑x
𝑑y 𝑑v 𝑑𝑢 1
∴ =𝑢 +𝑣 = x1/2(4 + 8x) + (1 + 2x)2 ( )
𝑑x 𝑑x 𝑑x 2x1/2
(1 + 2x)2
= x1/2(4 + 8x) +
2x1/2
2x(4 + 8x) + (1 + 2x)2
=
2x1/2
(d) 𝑦 = (5x4 + 1)7𝑠𝑖𝑛3x

Let 𝑢 = (5x4 + 1)7, 𝑣 = 𝑠𝑖𝑛3x

𝑑𝑢 𝑑𝑣
= 7(5x4 + 1)6(20x3) , = 3𝑐𝑜𝑠3x
𝑑x 𝑑x
𝑑y 𝑑v 𝑑𝑢
∴ =𝑢 +𝑣 = (5x4 + 1)7(3𝑐𝑜𝑠3x) + 𝑠𝑖𝑛3x(140x3(5x4 + 1)6)
𝑑x 𝑑x 𝑑x
= 3𝑐𝑜𝑠3x(5x4 + 1)7 + 140x3𝑠𝑖𝑛3x(5x4 + 1)6

3. Derivative of a Quotient.
𝑢
If 𝑦 = where 𝑢 and 𝑣 are functions of x such that 𝑣 ≠ 0 then the derivative of 𝑦
𝑣
with respect to x is
𝑑𝑢 𝑑𝑣
𝑑y 𝑣 𝑑x − 𝑢 𝑑x
=
𝑑x 𝑣2
Example3: Find the derivative of each of the following.

1 + x2 3 + 2x − x2
(a) 𝑦 = (𝑏) 𝑦 =
1 − x2 √1 + x
Calculus prepared b Engr. James

2+x 3√(1 + 3x2)2

(𝑐) 𝑦 = (𝑑) 𝑦=
x2 + 2x + 7 x
Solution
1 + x2
(a) 𝑦 =
1 − x2
Let 𝑢 = 1 + x2, 𝑣 = 1 − x2
𝑑𝑢 𝑑𝑣
= 2x , = −2x
𝑑x 𝑑x
𝑑𝑢 𝑑𝑣
𝑑y 𝑣 −𝑢
𝑑x = (1 − x )2x − (1 + x )(−2x)
2 2
∴ = 𝑑x
𝑑x 𝑣2 (1 − x2)2
(2x − 2x3) + (2x + 2x3) 4x
= =
(1 − x2)2 (1 − x2)2
3 + 2x − x2
(b) 𝑦 =
√1 + x

Let 𝑢 = 3 + 2x − x2, 𝑣 = √1 + x = (1 + x)1/2

𝑑𝑢 𝑑𝑣 1
= 2 − 2x , =
𝑑x 𝑑x 2(1 + x)1/2
1
𝑑𝑢 𝑑𝑣 ((1 + x)1/2)2(1 − x) − (3 + 2x − x2) ( )
𝑑y 𝑣 𝑑x − 𝑢 𝑑x 2(1 + x) 1/2
∴ = =
𝑑x 𝑣2 ((1 + x) )
1/2 2

3 + 2x − x2
2((1 + x )1/2 )(1 − x) − ( )
= 2(1 + x)1/2
1+x
4(1 + x)(1 − x) − (3 + 2x − x2)
2(1 + x)1/2
=
1+x
4 − 4x2 − 3 − 2x + x2
=
2(1 + x)3/2
Calculus prepared b Engr. James

1 − 2x − 3x2 1 − 2x − 3x2
= =
2(1 + x)3/2 2 (√(1 + x)3)

2+x
𝑦=
(c) x2 + 2x + 7

Let 𝑢 = 2 + x, 𝑣 = x2 + 2x + 7
𝑑𝑢 𝑑𝑣
=1, = 2x + 2
𝑑x 𝑑x
𝑑𝑢 𝑑𝑣
𝑑y 𝑣 −𝑢
𝑑x = (x + 2x + 7) − (2 + x)(2x + 2)
2
∴ = 𝑑x
𝑑x 𝑣2 (x2 + 2x + 7)2
x2 + 2x + 7 − 2x2 − 6x − 4
=
(x2 + 2x + 7)2
−x2 − 4x + 3 3 − 4x − x2
= =
(x2 + 2x + 7)2 (x2 + 2x + 7)2

3√(1 + 3x2)2
(d) 𝑦 =
x
3
Let 𝑢 = √(1 + 3x2)2 = (1 + 3x2)2/3, 𝑣=x
𝑑𝑢 2 4x 𝑑𝑣
= (1 + 3x2)−1/36x = , =1
𝑑x 3 (1 + 3x2)1/3 𝑑x
𝑑𝑢 4x2 𝑑𝑣 ( 2)2/3
𝑑y 𝑣 𝑑x − 𝑢 𝑑x (1 + 3x2)1/3 − 1 + 3x
∴ = =
𝑑x 𝑣2 x2
4x2 − 1 − 3x2
(1 + 3x2)1/3 x2 − 1
= =
x2 x2(1 + 3x2)1/3
x2 − 1
=
3
x2 ( √(1 + 3x2))
Calculus prepared b Engr. James

Implicit Differentiation

The process of differentiating implicit function is called implicit differentiation.

Example4: Differentiate the following implicitly.

(a) x2 + 𝑦2 = 4 (𝑏) x2𝑦 + xy2 + 4x = 1

Solution

(a) x2 + 𝑦2 = 4

Differentiating term by term with respect to x we have

𝑑y
2x + 2y =0
𝑑x
𝑑y
2y = −2x
𝑑x
𝑑y −2x
=
𝑑x 2y
𝑑y −x
=
𝑑x y

(b) x2𝑦 + xy2 + 4x = 1

𝑑y 𝑑y
2xy + x2 + y2 + 2xy +4 =0
𝑑x 𝑑x
𝑑y 𝑑y
x2 + 2xy = −2xy − y2 − 4
𝑑x 𝑑x
𝑑y
(x2 + 2xy) = −2xy − y2 − 4
𝑑x
𝑑y −2xy − y2 − 4
=
𝑑x x2 + 2xy

Assignment

(𝑎) 4xy2 − 5x2y3 + 4𝑦 = 0 (𝑏) (x + 𝑦)2 = 5 (𝑐) 3x2 + 3xy2 − 𝑐𝑜𝑠2𝑦 = 0

Calculus prepared b Engr. James

Parametric Equation

Given that x = sint and y = cost where t is a parameter are called parametric
equation.

Example5: Given that x = 5t3 and y = 4t2 , find 𝑑y

𝑑x

Solution

x = 5t3
𝑑x 𝑑t 1
= 15t2 → =
𝑑t 𝑑x 15t2
y = 4t2
𝑑y
= 8t
𝑑t
𝑑y 𝑑y 𝑑𝑡 1 8
∴ = × = 8t × =
𝑑x 𝑑t 𝑑x 15t2 15t
Higher Derivatives
𝑑y
Given that y = 𝑓(x), is also a function of x. The derivative of 𝑑y
with respect
𝑑x 𝑑x
𝑑y 𝑑y
to x is 𝑑 ( ), 𝑑 ( ) is called the second derivative of 𝑦 with respect to x and is
𝑑x 𝑑x 𝑑x 𝑑x 2y
usually denoted with 𝑑 .
𝑑x2
3y 4y 𝑛y
The third derivative is 𝑑 while the fourth derivative is 𝑑 . In general 𝑑 is the
𝑑x3 𝑑x4 𝑑x𝑛
𝑛𝑡ℎ derivative of 𝑦 with respect to x.

Example6: Find the first, second and third derivatives of the following.
4
(a) 𝑦 = 3x4 (𝑏) 𝑦 = 𝑒x (𝑐) 𝑦 = 𝑙𝑛x (𝑑) 𝑦 = 𝑠𝑖𝑛3x

Solution

(a) 𝑦 = 3x4
𝑑y
= 12x3
𝑑x
Calculus prepared b Engr. James

𝑑2y
= 36x2
𝑑x2

𝑑3y
= 72x
𝑑x3
4
(b) 𝑦 = 𝑒x
𝑑y 4
= 4x3𝑒x
𝑑x
𝑑2y 4 4 4 4
= 12x2. 𝑒x + 4x3. 4x3𝑒x = 12x2𝑒x + 16x6𝑒x
𝑑x2
𝑑3y 4 4 4 4
= 24x. 𝑒x + 12x2. 4x3𝑒x + 96x5. 𝑒x + 16x6. 4x3𝑒x
𝑑x3
4 4 4 4
= 24x𝑒x + 48x5𝑒x + 96x5𝑒x + 64x9𝑒x

(𝑐) 𝑦 = 𝑙𝑛x
𝑑y 1
=
𝑑x x
𝑑2y −1
=
𝑑x2 x2
𝑑3y 2
=
𝑑x3 x3
Assignment

1. Find the first, second and third derivatives of 𝑦 = 𝑠𝑖𝑛3x.

2. Given that 𝑦 = 𝑎𝑐𝑜𝑠𝑘x + 𝑏𝑠𝑖𝑛𝑘x, show that
𝑑2y
+ k2𝑦 = 0
𝑑x2
3. Given that 𝑦 = 𝑒x − 𝑒−x, show that
𝑑3y 𝑑2y 𝑑y
+ 2+ + 𝑦 = 4𝑒x
𝑑x3 𝑑x 𝑑x
Calculus prepared b Engr. James

Applications of Differential Calculus

1. Tangents and Normals to Curves.

At any point on a curve, 𝑑y at that point gives the gradient of the tangent at the
𝑑x
point. A straight line perpendicular to the tangent at the point of contact of the
tangent to the curve is called a normal to the curve.

Example1: Find the equation of the tangent and the normal to the curve

𝑦 = 2x3 − x2 + 3x + 1 at the point x = 1.

Solution

𝑦 = 2x3 − x2 + 3x + 1
𝑑y
= 6x2 − 2x + 3
𝑑x
𝑑y
= 6(1)2 − 2(1) + 3 = 7
𝑑x x=1

If 𝑚 is the gradient of the tangent at the point x = 1, then 𝑚 = 7

At the point x = 1, we have that 𝑦 = 2 − 1 + 3 + 1 = 5

The equation of the tangent at the point x = 1 is given by

y − y1 = 𝑚(x − x1)

y − 5 = 7(x − 1)

y − 5 = 7x − 7

y − 7x + 2 = 0

If 𝑚′ is the gradient of the normal at x = 1 , then

−1 −1
𝑚′ = =
𝑚 7
Hence, the equation of the normal at the point x = 1 is
−1
y − y1 = (x − x1)
𝑚
Calculus prepared b Engr. James

−1
y−5 = (x − 1)
7
7(y − 5) = −1(x − 1)

7𝑦 − 35 = 1 − x

x + 7𝑦 − 36 = 0

Example2: Find the equation of the tangent to the curve x2𝑦 + xy3 + 3x − 13 = 0

At the point (1, 2)

Solution

x2𝑦 + xy3 + 3x − 13 = 0
𝑑y 𝑑y
2x𝑦 + x2 + y3 + 3xy2 +3 =0
𝑑x 𝑑x
𝑑y 𝑑y
x2 + 3xy2 = −2x𝑦 − y3 − 3
𝑑x 𝑑x
𝑑y
(x2 + 3xy2) = −2x𝑦 − y3 − 3
𝑑x
𝑑y −2x𝑦 − y3 − 3
=
𝑑x x2 + 3xy2
𝑑y −2x𝑦 − y3 − 3 −4 − 8 − 3 −15
= = =
𝑑x x=1, 𝑦=2 x2 + 3xy2 1 + 12 13
−15
If 𝑚 is the gradient of the tangent at the point x = 1, 𝑦 = 2, then 𝑚 = .
13
Therefore the equation of the tangent at the point x = 1, 𝑦 = 2 is given by

y − y1 = 𝑚(x − x1)
−15
y−2= (x − 1)
13
13(y − 2) = −15(x − 1)

13y − 26 = −15x + 15
Calculus prepared b Engr. James

15x + 13y = 41

15x + 13y − 41 = 0

2. Maximum and Minimum Points.

𝑌

𝑓′(x) = 0
𝐴
𝑓(x) = +𝑣𝑒 𝑓(x) = −𝑣𝑒 𝑓(x) = +𝑣𝑒

𝑓′(x) = 0
𝑋

The point A in the figure above where the gradient is changing from positive
through zero to negative is called the maximum point. The point B where the
gradient is changing from negative through zero to positive is called a minimum
point. The maximum and the minimum points are both called turning points. The
point where 𝑓′(x) = 0 is called a stationary point. The value of 𝑦 at the maximum
point is called the maximum value while the minimum point of 𝑦 is called the
minimum value.
To find the turning points of a given function, we shall carry out the following test

𝑑y 𝑑2y
(𝑎) = 0, < 0 (Maximum point)
𝑑x 𝑑x2
𝑑y 𝑑2y
(𝑏) = 0, > 0 (Minimum point)
𝑑x 𝑑x2
Calculus prepared b Engr. James

Example1: Find and classify the turning points of the following curves

1. 𝑦 = x2 − 5x + 6
2. 𝑦 = x2 + 4x − 3

Solution

1. 𝑦 = x2 − 5x + 6
𝑑y
= 2x − 5
𝑑x
𝑑y
=0
𝑑x
2x − 5 = 0
5
x=
2
𝑑2y
=2
𝑑x2
This shows that it is minimum point.

2. 𝑦 = x2 + 4x − 3
𝑑y
= 2x + 4
𝑑x
𝑑y
=0
𝑑x
2x + 4 = 0

x = −2
𝑑2y
=2
𝑑x2
This shows that it is minimum point.
Calculus prepared b Engr. James

Assignment

Find the turning points of the following curve 𝑦 = x3 − 12x + 6.

Review of Integration
Differentiation measures the rate of change while integration measures area.
If 𝑑y = 𝑓(x), then 𝑑y = 𝑑x𝑓(x)
𝑑x

∫ 𝑑y = ∫ 𝑑x𝑓(x)

y = 𝐹(x)x + C

where C is called an arbitrary constant of integration.

Example1: Evaluate the following integrals

(𝑎) ∫(x 7 + 1)x (𝑏) ∫ √x𝑑x (𝑐) ∫(2x2 + 3x + 8)𝑑x

Solution

(𝑎) ∫(x 7 + 1)x = ∫ x7𝑑x + ∫ 1𝑑x

x8
= +x+C
8

(𝑏) ∫ √x𝑑x = ∫ x1/2𝑑x

2x3/2
= +C
3

(𝑐) ∫(2x2 + 3x + 8)𝑑x = ∫ 2x2𝑑x + ∫ 3x𝑑x + ∫ 8𝑑x

2x3 3x2
= + + 8x + C
3 2
Techniques of Integration

1. The Integral of the form

Calculus prepared b Engr. James

∫ (𝑎x + 𝑏)x
Let 𝑈 = 𝑎x + 𝑏
𝑑u
=𝑎
𝑑x
𝑑x 1
=
𝑑u 𝑎
1
𝑑x = 𝑑u
𝑎
1 1
∴ ∫ (𝑎x + 𝑏)x = ∫ 𝑓(𝑢) 𝑑𝑢 = ∫ 𝑓(𝑢)𝑑𝑢.
𝑎 𝑎
Example2: Evaluate the following integrals

3𝑑x
(𝑎) ∫(2x + 3)5𝑑x (𝑏) ∫ (𝑐) ∫ 𝑒−5x+2𝑑x
(4x − 1)3

Solution

(𝑎) ∫(2x + 3)5𝑑x

Let 𝑈 = 2x + 3
𝑑u
=2
𝑑x
𝑑x 1
=
𝑑u 2
1
𝑑x = 𝑑u
2
1
∫(2x + 3)5𝑑x = ∫ 𝑈5 𝑑u
2
1 1 6 1
= ∫ 𝑈5𝑑u = 𝑈 +𝐶 = (2x + 3)6 + 𝐶
2 12 12
Calculus prepared b Engr. James

3𝑑x
(𝑏) ∫
(4x − 1)3
Let 𝑈 = 4x − 1
𝑑u
=4
𝑑x
𝑑x 1
=
𝑑u 4
1
𝑑x = 𝑑u
4
3𝑑x 3 1
∫ = ∫ 𝑑u
(4x − 1) 3 𝑈3 4

3 1 3 3 𝑈−2 −3 −3
= ∫ 𝑑u = ∫ 𝑈−3𝑑u = +𝐶 = +𝐶 = +𝐶
4 𝑈3 4 4 −2 8𝑈2 8(4x − 1)2
(𝑐) ∫ 𝑒−5x+2𝑑x

Let 𝑈 = −5x + 2
𝑑u
= −5
𝑑x
𝑑x −1
=
𝑑u 5
−1
𝑑x = 𝑑u
5
−1 −1 −1 1
∫ 𝑒−5x+2𝑑x = ∫ 𝑒𝑢 𝑑u = ∫ 𝑒𝑢𝑑u = 𝑒𝑢 + 𝐶 = − 𝑒−5x+2 + 𝐶
5 5 5 5
2. The Integral of the form
𝑓′(x)
∫ 𝑑x
𝑓(x)

Let 𝑈 = 𝑓(x)
Calculus prepared b Engr. James

𝑑u
= 𝑓′(x)
𝑑x
𝑑u = 𝑓′(x)x
𝑑u 𝑓′(x)
= 𝑑x
𝑢 𝑓(x)
𝑑𝑢 𝑓′(x)
∫ =∫ 𝑑x
𝑢 𝑓(x)
𝑑𝑢
∫ = ∫ 𝑢−1𝑑𝑢 = 𝐿𝑜𝑔 𝑢.
𝑒
𝑢
𝑓′(x)
𝐿𝑜𝑔𝑒𝑢 = ∫ 𝑑x
𝑓(x)
𝑓′(x)
Hence, ∫ 𝑑x = 𝐿𝑜𝑔𝑒𝑓(x) + C
𝑓(x)
Example3: Evaluate the following integrals
2x𝑑x 3x2 + 4x
(𝑎) ∫ 𝑑x (𝑏) ∫ 𝑑x
x2 + 1 x3 + 2x2 + 7
Solution
2x𝑑x
(𝑎) ∫ 𝑑x
x2 + 1
𝑑 2
(x + 1) = 2x
𝑑x
2x𝑑x (x2 + 1)
∴∫ 2 𝑑x = ∫ 𝑑x 2 𝑑x = 𝐿𝑜𝑔𝑒(x2 + 1) + 𝐶
x +1 x +1
3x2 + 4x
(𝑏) ∫ 𝑑x
x3 + 2x2 + 7
𝑑
(x3 + 2x2 + 7) = 3x2 + 4x
𝑑x
Calculus prepared b Engr. James

3x2 + 4x (x3 + 2x2 + 7)

∴∫ 3 𝑑x = ∫ 𝑑x 3 𝑑x = 𝐿𝑜𝑔𝑒 (x3 + 2x2 + 7) + 𝐶
x + 2x + 7
2
(x + 2x2 + 7)

3. Integration by algebraic substitution

An integral that is not in the standard form can be reduced to one that is in the
standard form by making appropriate algebraic substitution
Example4: Evaluate the following integrals

2
(𝑎) ∫ x𝑒x 𝑑x (𝑏) ∫ x2𝑐𝑜𝑠x3𝑑x

Solution
2
(𝑎) ∫ x𝑒x 𝑑x

Let 𝑈 = x2
𝑑u
= 2x
𝑑x
𝑑x 1
=
𝑑u 2x
1
𝑑x = 𝑑u
2x
2 1 1 1 1 2
𝑢
∴ ∫ x𝑒x 𝑑x = ∫ x
𝑒 𝑑u = ∫ 𝑒 𝑢𝑑u = 𝑒𝑢 + 𝐶 = 𝑒x + 𝐶
2x 2 2 2

(𝑏) ∫ x2𝑐𝑜𝑠x3𝑑x

Let 𝑈 = x3
𝑑u
= 3x2
𝑑x
𝑑x 1
=
𝑑u 3x2
Calculus prepared b Engr. James

1
𝑑x = 𝑑u
3x2
1 1
∴ ∫ x2𝑐𝑜𝑠x3𝑑x = ∫ x2𝑐𝑜 1 1 3
𝑑u = ∫ 𝑐𝑜𝑠𝑢𝑑u = 𝑠𝑖𝑛𝑢 + 𝐶 = 𝑠𝑖𝑛x + 𝐶
3x2 3 3 3
Assignment

Evaluate the following integrals

(𝑎) ∫ x√1 + x2𝑑x (𝑏) ∫ x2(3 + x3)3/2𝑑x

4. Integration by Parts
This is a technique use for integrating product were substitution method fails. If
𝑦 = uv, then by product rule we have
𝑑y 𝑑v 𝑑𝑢
=𝑢 +𝑣
𝑑x 𝑑x 𝑑x
𝑑y 𝑑v 𝑑𝑢
∴ ∫ ( ) 𝑑x = ∫ (𝑢 ) 𝑑x + ∫ (𝑣 ) 𝑑x
𝑑x 𝑑x 𝑑x

∴ 𝑦 = ∫ 𝑢𝑑v + ∫ 𝑣𝑑𝑢

∴ ∫ 𝑢𝑑v = 𝑦 − ∫ 𝑣𝑑𝑢

∴ ∫ 𝑢𝑑v = uv − ∫ 𝑣𝑑𝑢

Example5: Evaluate the following integrals

(𝑎) ∫ x𝑠𝑖𝑛x𝑑x (𝑏) ∫ x2𝑐𝑜𝑠x𝑑x

Solution

(𝑎) ∫ x𝑠𝑖𝑛x𝑑x

Let 𝑢 = x
Calculus prepared b Engr. James

𝑑u
=1
𝑑x
𝑑𝑣 = 𝑠𝑖𝑛xdx

𝑣 = −𝑐𝑜𝑠x

∴ ∫ 𝑢𝑑v = uv − ∫ 𝑣𝑑𝑢

∫ x𝑠𝑖𝑛x𝑑x = −xcosx − ∫ cosxdx

∫ x𝑠𝑖𝑛x𝑑x = sinx − xcosx + C

(𝑏) ∫ x2𝑐𝑜𝑠x𝑑x

Let 𝑢 = x2
𝑑u
= 2x
𝑑x
𝑑𝑣 = 𝑐𝑜𝑠xdx

𝑣 = 𝑠𝑖𝑛x

∴ ∫ 𝑢𝑑v = uv − ∫ 𝑣𝑑𝑢

∫ x𝑠𝑖𝑛x𝑑x = x2sinx − ∫ 2xsinxdx

= x2sinx − 2 ∫ xsinxdx

= x2sinx − 2sinx + 2xcosx + C

Assignment

Evaluate the following integrals

(𝑎) ∫ x2lnx𝑑x (𝑏) ∫ 𝑒2x𝑐𝑜𝑠2x𝑑x

Calculus prepared b Engr. James

5. Integration by Partial Fractions

If a rational expression is not in a standard integral form, it could be transformed
into a standard form by splitting it into partial fractions.

Example6: Evaluate the following integrals

4x − 5 2x3 − 2x2 − 2x − 7
(𝑎) ∫ 𝑑x (𝑏) ∫ 𝑑x
(x + 1)(x − 2) x2 − x − 2

Solution
4x − 5
First split into its partial fractions.
(x + 1)(x − 2)
4x − 5 A B
= +
(x + 1)(x − 2) (x + 1) (x − 2)

A(x − 2) + B(x + 1) = 4x − 5

Using cover up method, to solve for A let x = −1 and to solve for B let x = 2

A(−1 − 2) = 4(−1) − 5

−3A = −9

A=3

Similarly, B = 1
4x − 5 3 1
∴ = +
(x + 1)(x − 2) (x + 1) (x − 2)
4x − 5 3 1
∫ 𝑑x = ∫ ( + ) 𝑑x
(x + 1)(x − 2) (x + 1) (x − 2)
3 1
=∫( ) 𝑑x + ∫ ( ) 𝑑x
x+1 x−2
= 3𝑙𝑛(x + 1) + 𝑙𝑛(x − 2) + C

2x3 − 2x2 − 2x − 7
(𝑏) ∫ 𝑑x
x2 − x − 2
Calculus prepared b Engr. James

Since the degree of the numerator is higher than the degree of the denominator, the
expression must be in a proper algebraic fraction form.
2x

x2 − x − 2 2x3 − 2x2 − 2x − 7
2x3 − 2x2 − 4x
2x − 7

2x3 − 2x2 − 2x − 7 2x − 7
∴ = 2x +
x2 − x − 2 x2 − x − 2
2x − 7
Splitting into partial fraction, we have
x2 − x − 2
2x − 7 A B
= +
x2 − x − 2 (x + 1) (x − 2)

A(x − 2) + B(x + 1) = 2x − 7

Solving for A and B, we have A = 3 and B = −1 respectively

2x − 7 3 1
= −
x2 − x − 2 (x + 1) (x − 2)

2x3 − 2x2 − 2x − 7 3 1
∴ = 2x + −
x2 − x − 2 (x + 1) (x − 2)

2x3 − 2x2 − 2x − 7 3 1
∴ ∫ dx = ∫ (2x + − ) dx
x2 − x − 2 (x + 1) (x − 2)

= x2 + 3ln(x + 1) − ln(x − 2) + C

Assignment

Evaluate the following integrals

4x − 23 2
(𝑎) ∫ 𝑑x (𝑏) ∫ 𝑑x
(x − 5)2 x2 − 4