Integration using partialfractions

CHAPTER 51

Learning Materials

Type of Partial

Fractions

1

Problems #1:

with linear

factors

2

Problems #2:

with repeated

linear factors

3

Problems #3:

with quadratic

factors

4

Introduction

A rational fraction is a quotient of two polynomials, say:

)(

)(

1

1

10

1

1

10

xP

xP

axaxaxa

bxbxbxb

n

m

nn

nn

mm

mm

=++++

++++

−

−

−

−

Suppose that Pm(x) dan Pn(x) have no common factor other than 1

Our Idea Solution :• We shall express again a rational function as a sum of simpler so-called

“Partial Fractions”

Example : Solve this problem +−

+dx

67xx

12x3

Introduction

• If the degree of Pm (x) is greater or equal to the degree of Pn (x), then is used long division of the polynomial to get the result P (x) and a remainder of R (x), so that it can be written as follows:

)(

)()(

)(

)(

xP

xRxP

xP

xP

nn

m +=

• where the degree of R (x) is smaller than the degree of Pn (x)

• Factorization of denominator Pn(x) = p1(x) p2(x) ... pk(x), where each factor has the form as follows:

• Linear factor :

• Repeated linear factor :

• Quadratic factor :

• Repeated quadratic factor :

α)(x −rα)(x −

s2 c)bx(ax ++

04acb dimana c)bx(ax 22 −++

Type of Partial Fractions

Problem #1: with linear factors

All the factors of the denominator linear and none repeated

Example :

1 .dx67xx

12x Calculate

3 +−

+

CαxlnA dxα-x

A Thus, +−=

First, we must to factoring of denominator: 67xx3 +−

3)2)(x-1)(x-(x67xx3 +=+−

3 x

C

2 -x

B

1 -x

A

3) -2)(x -1)(x -(x

12x

+++=

+

Problem #1: with linear factors

Thus,

3) 2)(x -1)(x -(x

2) -1)(x -C(x 3)1)(x -B(x 3) 2)(x -A(x

3) 2)(x -1)(x -(x

12x

+

++++=

+

+

2) -1)(x -C(x 3)1)(x -B(x 3) 2)(x -A(x 12x ++++=+

Coefficient x2→ 0 = A + B + C

Coefficient x → 2 = A + 2B – 3C

Coefficient x0→ 1 = -6A – 3B + 2C

So that it is obtained the value of A = - ¾ ; B = 1 ; and C = - ¼

Problem #1: with linear factors

Thus,

+−

−+

−−=

+−

+

3x

dx

4

1

2x

dx

1x

dx

4

3dx

67xx

12x3

C3xln4

12xln1xln

4

3++−−+−−=

( )C

3)(x1)(x

2xln

4

13

4

++−

−=

Exercise 1

Problem #2: with repeated linear factors

All the factors of the denominator linear but some repeated

Cα)1)(x(r

R-dx

α)-(x

R Thus,

1 -r r+

−−=

Example :

2 .dx)1)(2(

124x Determine

3

23

+−

++−

xx

xx

323

23

)1()1(12)1)(2(

124x

++

++

++

−=

+−

++−

x

D

x

C

x

B

x

A

xx

xx

2)D(x1)2)(xC(x

1)2)(xB(x1)A(x1x2x4x 2323

−++−

++−++=++−

Problem #2: with repeated linear factors

Thus : Coefficient x3→ 4 = A + B

Coefficient x2→ -2 = 3A + C

Coefficient x → 1 = 3A - 3B – C + D

Coefficient x0→ 1 = A – 2B - 2C – 2D

Then it is obtained the value of A = 1 ; B = 3 ; C = - 5 ; and D = 2

++

+−

++

−=

+−

++−323

23

)1(2

1)(x

dx5

1x

dx3

2x

dxdx

)1)(2(

124x

x

dx

xx

xx

C)1(

1

1

51xln32xln

2+

+−

++++−=

xx

C)1(

451)2)(x(xln

2

3 ++

+++−=

x

x

Exercise 2

Problem #3: with quadratic factors

Problem #3: with quadratic factors

Exercise

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