KNF 1013 Engineering Mathematics I

Today’s Objectives:
KNF 1023 At the end of the class, students would be able to:
ENGINEERING MATHEMATICS II
 Integrate functions by first expressing them in partial

METHOD OF INTEGRATION :- fractions.

PARTIAL FRACTION

 Integrate products of functions using integration by
 INTEGRATION BY PARTS parts.
 INTEGRATION BY SUBSTITUTION
 Integrate functions by making a substitution.

1 2

PARTIAL FRACTION PARTIAL FRACTIONS (continued)

 Integrate the followings :-

 This process of taking a rational expression and
6 x  13 x 5 1
dx;  x  32 dx;
 x 2  5x  6  x x 2
1  dx; decomposing it into simpler rational expressions that
we can add or subtract to get the original rational
 If the integrand is in the form of an algebraic fraction
expression is called partial fraction decomposition.
and the integral cannot be evaluated by simple
 Many integrals involving rational expressions can be
methods, the fraction needs to be expressed in partial
done if we first do partial fractions on the integrand.
fractions before integration takes place.

3 4

PARTIAL FRACTIONS (continued) PARTIAL FRACTIONS (continued)

Expressing fractional functions in partial fractions :- Example

6 x  13
Denominator containing… Expression Form of Partial Fractions
x 2
dx
 5x  6
Linear factor
f x A B

 x  a x  b   x  a  x  b  6 x  13 6 x  13 A B
  
x 2  5 x  6  x  2 x  3 x  2 x  3
Repeated linear factors Find the values of A & B by multiplying both sides :
f x  A B
 6 x  13  A x  3  B x  2 
 x  a 2  x  a   x  a 2
6 x  13   A  B x  3 A  2 B 
Quadratic term Equate x & constant terms :
(which cannot be factored) f x Ax  B
ax 2
 bx  c  ax 2  bx  c The x terms : - 6 x  13  A  B
The constant terms : - 13  3 A  2 B
5 6

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 KNF 1013 Engineering Mathematics I

PARTIAL FRACTIONS (continued)

PARTIAL FRACTIONS (continued)

Example
Solve for A & B : x5
 x  3 2
dx
A 1 & B  5
6 x  13 1 5
   x5 A B
x  2 x  3 x2 x3  
x  32 x  3  x  32
6 x  13 1 5
Thus, x 2
dx   dx   dx Multiply both sides by  x  3 :
2
 5x  6 x2 x3
 ln x  2   5 ln x  3  c x  5  A x  3  B
5 Equate x & constant terms :
 ln x  2 x  3  c
# The x terms : - 1  A
The constant terms : - 5  3 A  B
B2
7 8

PARTIAL FRACTIONS (continued)

PARTIAL FRACTIONS (continued)

Example
x5 1 2 1
  
x  32 x  3 x  32  xx 2
1  dx
x5 1 2
Thus,  dx   dx   dx 1 A Bx  C
x  32 x3 x  32   

x x2  1 x x2 1  
2
 ln x  3  c setting the numerators :
x  3
#  
1  A x 2  1   Bx  C x
Expand terms :
1   A  B x 2  Cx  A

9 10

PARTIAL FRACTIONS (continued)

equal the terms : -

PARTIAL FRACTIONS (continued)

x2 :  A B  0  B  A
x1 :  C0 Practice
x0 :  A 1  B  1
6
1
1 x a) x dx
   2
 2x  8

x x 1 x x2  1
2
  
1 1 x
Thus,  dx   dx   2 dx 9
x x2  1  x 
x 1 b)  x 1x  2 2
dx
1
 ln x  ln x 2  1  c
2
x
 ln c
x2 1
# 11 12

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 KNF 1013 Engineering Mathematics I

INTEGRATION BY PARTS (continued)

INTEGRATION BY PARTS
 The product rule for differentiation is
 Sometimes we meet an integration that is the product d du dv
(uv)  v u
dx dx dx
of two functions.
dv d du
2 3x  Rearrange the above rule u  (uv)  v
 x cos xdx; x e dx;  x ln x dx; dx dx dx

 Such products may be able to be integrate by using dv d uv  du

 Integrate both sides  u dxdx   dx   v dx
dx dx
Integration by Parts.
 Simplify dv du
 u dxdx  uv   v dx dx
(integration by parts formula)
13 14

INTEGRATION BY PARTS (continued)

INTEGRATION BY PARTS (continued)

Example
Example
 x ln x dx
 x cos dx dv
Let u  ln x and x
dx
dv
Let ux and  cos x du 1 x2
dx so,  v   xdx 
du dx x 2
so, 1 v   cos xdx  sin x Using the formula :
dx
Using the formula : dv du x2 x2 1
dv du
 u dxdx  uv   v dx dx   x ln x dx  ln x 
2  2 x
. dx

 u dxdx  uv   v dx dx x2 x
 ln x   dx
2 2
 x cos xdx  x sin x   sin x.1dx x2 x2
 x sin x  cos x  c  ln x   c
# 15 2 4 #
16

INTEGRATION BY PARTS (continued) INTEGRATION BY PARTS (continued)

Example Example
2 3x 2 3x
x e dx x e dx
dv
dv Let u  x2 and  e3 x
Let u  x2 and  e3 x dx
dx
du 1
du 1 so,  2x v   e3 x dx  e3 x
so,  2x v   e3 x dx  e3 x dx 3
dx 3
Using the formula :
Using the formula :
dv du 2 3x 1 1
dv du 2 3x 1 1  u dxdx  uv   v dx dx  x e dx  x 2 . e3 x   e3 x .2 xdx
 u dxdx  uv   v dx dx  x e dx  x 2 . e3 x   e3 x .2 xdx 3 3
3 3
1 2
1 2  x 2 e3 x   xe3 x dx
 x 2 e3 x   xe3 x dx 3 3
3 3
the resulting integral is still a product… do another by parts
# 17 18

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 KNF 1013 Engineering Mathematics I

INTEGRATION BY PARTS (continued) INTEGRATION BY PARTS (continued)

2 dv
Let u x and  e3 x
3 dx Practice
du 2 1
so,  v   e3 x dx  e3 x
dx 3 3
2 3x 1 2 3x 2 3x a)  ln xdx
 x e dx  x e  3 xe dx
3
1 2 3x  2 1 3 x 1 2 
 x e   x. e   e3 x . dx b)
3 3 3 3 3   x sin 2 xdx
1 2 2
 x 2 e 3 x  xe3 x  e3 x  c
3 9 27
#
19 20

INTEGRATION BY SUBSTITUTION INTEGRATION BY SUBSTITUTION

(continued)

Substituting u = ax + b Example

 Has been done before...  cos3x  4 dx

take u  3 x  4
Example
du 1
5 3  dx  du
 x  4 dx dx 3
take u  x  4 1 
 cos 3 x  4 dx   u 3 du 
cos
du
1  du  dx
dx 1
5
  cos udu
 x  4 dx   u du
5
3
1 1
u6
6
 x  4  c  sin u  c  sin 3x  4  c
 c  3 3
6 6 # 21 # 22

INTEGRATION BY SUBSTITUTION INTEGRATION BY SUBSTITUTION

(continued) (continued)
Example
Substituting u = g(x)
1
 1  2 x dx  The composite-function rule for differentiation
take u  1  2 x d
du 1  f ( g ( x))  f ' ( g ( x)) g ' ( x )
 2  dx   du dx
dx 2 can be used to evaluate some integrals.
1 1 1   Reversing the differentiation process gives,
 1 2 x dx  
u 2 
 du 

1 1
   du
 f ' g x g ' x dx  f g x   c
2 u
 The key step here is to identify the function g(x).
1 1
  ln u  c   ln 1  2 x  c Different choices of g(x) may provide answers that
2 2
# 23
differ by a constant. 24

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 KNF 1013 Engineering Mathematics I

INTEGRATION BY SUBSTITUTION INTEGRATION BY SUBSTITUTION

(continued) (continued)

 To make the process of manipulation easier to follow, Example

it is usual to set u = g(x), so that the integral becomes
 2x ( x 2  3)dx
du
 f ' g ( x) g ' ( x) dx   f ' (u ) dx
dx take u  x 2  3
  f ' (u ) du du
 2x
 f (u )  c dx
 du 
 Substituting back t = g(x) into the equation gives,  2x ( x 2  3)dx     u dx
 dx 
 
 f '  g ( x)g ' ( x)dx  f g ( x)  c 1/ 2
  u  du
This technique for evaluating integrals is called the 2 3/ 2 2

 
3/ 2
u  c  x2  3  c 
substitution method. 3 3 #
25 26

INTEGRATION BY SUBSTITUTION
(continued) INTEGRATION BY SUBSTITUTION
Example (continued)

4x Example
 2x 2  1
dx
 cos3x 12x dx
4 3

2
take u  2 x  1
take u  3x 4
du
 4x  4 xdx  du du
dx  12 x 3  12 x 3 dx  du
dx
4x 1
 cos3x 12x dx   cos u .du
 dx   du 4 3
2x 2 1 u
  u 
1/ 2
du  sin u  c
 
 sin 3x 4  c
 2u1/ 2  c  2 2 x 2  1  c #
#
27 28

INTEGRATION BY SUBSTITUTION
(continued)

Practice

a)  2 xe
x2
dx
THANK YOU
2
b)  5x 1  x 3 dx

29 30

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