Download PPT 6.3 Partial Fractions

6.3 Partial Fractions
Rational Functions
Definition
Example
A function of the type P/Q, where both P and Q
are polynomials, is a rational function.
x3  1
is a rational function.
2
x  x 1
The degree of the denominator of the above rational function is less than the
degree of the numerator. First we need to rewrite the above rational
function in a simpler form by performing polynomial division.
Rewriting
x3  1
2

x

1

x2  x  1
x2  x  1
For integration, it is always necessary to perform polynomial division first, if
possible. To integrate the polynomial part is easy, and one can reduce the
problem of integrating a general rational function to a problem of integrating a
rational function whose denominator has degree greater than that of the
numerator (is called proper rational function). Thus, polynomial division is the
first step when integrating rational functions.
Partial Fraction Decomposition
The second step is to factor the denominator Q(x) as far as
possible. It can be shown that any polynomial Q can be factored as
a product of linear factors (of the form ax+b) and irreducible
quadratic factors (of the form ax2+bx+c, where b2-4ac<0). For
instance, if Q(x)=x4-16 then Q(x) = (x2-4)(x2+4)=(x-2)(x+2)(x2+4)
The third step is to express the proper rational function as a sum of
partial fractions of the form
A / (ax+b)i
Example:
or
(Ax+b) / (ax2+bx+c)j
3x 2  3x  2
1
1
2x



x 3  x 2  x  1 1 x 1 x 2 1 x 2
The fourth step is to integrate the partial fractions.
Integration Algorithm
Integration of a rational function f = P/Q, where P and Q
are polynomials can be performed as follows.
1. If deg(Q)  deg(P), perform polynomial division and write
P/Q = S + R/Q, where S and R are polynomials with
deg(R) < deg(Q).
Integrate the polynomial S.
2. Factorize the polynomial Q.
3. Perform Partial Fraction Decomposition of R/Q.
4. Integrate the Partial Fraction Decomposition.
Different cases of Partial Fraction
Decomposition
The partial fraction decomposition of a rational function R=P/Q,
with deg(P) < deg(Q), depends on the factors of the denominator Q.
It may have following types of factors:
1. Simple, non-repeated linear factors ax + b.
2. Repeated linear factors of the form (ax + b)k, k > 1.
3. Simple, non-repeated quadratic factors of the type ax2 + bx + c.
Since we assume that these factors cannot anymore be factorized,
we have b2 – 4 ac < 0.
4. Repeated quadratic factors (ax2 + bx + c)k, k>1. Also in this case
we have b2 – 4 ac < 0.
We will consider each of these four cases separately.
Simple Linear Factors
Case I
Consider a rational function of the type
Px
Qx

Px
 a1x  b1  a2 x  b2   an x  bn 
bi b j
where a j  0 for all j,

for i  j, and deg P   n  deg Q .
ai a j
Partial Fraction Decomposition: Case I
Px
 a1x  b1  a2 x  b2  an x  bn 

A1
A2


a1x  b1 a2 x  b2
for some uniquely defined numbers Ak , k  1, , n.

An
an x  bn
Simple Linear Factors
Example
Consider the rational function
2
2

.
2
x  1  x  1 x  1
By the result concerning Case I we can find numbers A and B such that
2
2
A
B



.
x 2  1  x  1 x  1 x  1 x  1
Compute these numbers in the following way
2
A
B
2
A( x  1)
B( x  1)





x2  1 x  1 x  1
x 2  1 ( x  1)( x  1) ( x  1)( x  1)
A  B  0
 A 1
0 x  2 ( A  B )x  ( A  B )
 2



.
x 1
x2  1
A

B

2
B


1


So the partial fraction decomposition is
To get the
equations for A
and B we use
the fact that two
polynomials are
the same if and
only if their
coefficients are
the same.
2
1
1


.
2
x 1 x 1 x 1
Simple Quadratic Factors
Case II
Consider a rational function of the type
Px
Qx
, deg P   deg  Q  .
Assume that the denominator Q  x  has a quadratic factor ax 2  bx  c.
Partial Fraction Decomposition: Case II
The quadratic factor ax 2  bx  c of the denominator leads to a term
of the type
Ax  B
in the partial fraction decomposition.
ax 2  bx  c
Simple Quadratic Factors
Example
The rational function
the type
3
3

has a term of
x 3  1 ( x  1)( x 2  x  1)
Ax  B
in its partial fraction decomposition.
x2  x  1
3
Ax  B
C
3
( Ax  B)( x  1)
C( x 2  x  1)


 3


x3  1 x2  x  1 x  1
x  1 ( x  1)( x 2  x  1) ( x 2  x  1)( x  1)
A C  0


3
( A  C )x 2  (C  B  A)x  C  B  C  B  A  0

3

x 1
x3  1
C B  3

Hence
3
1
x2


x3  1 x  1 x 2  x  1
 A  1

 B  2.
 C 1

To get these equations use the fact
that the coefficients of the two
numerators must be the same.
Repeated Linear Factors
Case III
Consider a rational function of the type
Px
Qx
, deg P   deg  Q  .
Assume that the denominator Q  x  has a repeated linear factor
 ax  b 
k
, k  1.
Partial Fraction Decomposition: Case III
The repeated linear factor
of the type
 ax  b 
A1
A2


2
ax  b  ax  b 
decomposition.

k
of the denominator leads to terms
Ak
 ax  b 
k
in the partial fraction
Repeated Linear Factors
Example
4x 2  4x  4
4x 2  4x  4
The rational function 3

has
2
2
x  x  x  1  x  1 x  1
a partial fraction decomposition of the type
4x 2  4x  4
A
B
C




x3  x2  x  1
x  1  x  12 x  1
A  x  1 x  1  B  x  1  C  x  1
 x  1 x  1
2
2
4x 2  4x  4
 3
x  x2  x  1
A  C  x 2   B  2C  x  A  B  C


2
 x  1 x  1
A  3

 B  2
C  1

A
B
C


.
x  1  x  12 x  1
Equate the
coefficients of the
numerators.
A C  4

4x 2  4x  4

 3

 B  2C  4
2
x  x  x 1
  A  B  C  4

4x 2  4x  4
3
2
1
We get 3



.
x  x2  x  1
x  1  x  12 x  1
Repeated Quadratic Factors
Case IV
Consider a rational function of the type
Px
Qx
, deg P   deg  Q  .
Assume that the denominator Q  x  has a repeated quadratic factor
 ax
2

k
 bx  c , k  1.
Partial Fraction Decomposition: Case IV
The repeated quadratic factor
to terms of the type

ax 2  bx  c

k
of the denominator leads
A1x  B1
A2 x  B2

ax 2  bx  c
ax 2  bx  c

in the partial fraction decomposition.

2


Ak x  Bk
 ax
2
 bx  c

k
Repeated Quadratic Factors
2x 4  3 x 2  x
2x 4  3 x 2  x

has a partial
2
5
4
3
2
2
x  x  2x  2x  x  1  x  1 x  1
Example

fraction decomposition of the type
A1x  B1 A2 x  B2
C



2
2
2
x 1
x 1
x 1


A1x  B1 A2 x  B2
C


.
2
2
2
x 1
x 1
x 1







 A1x  B1  x 2  1  x  1   A2 x  B2  x  1  C x 2  1
 x  1  x
2

1
2
2
Computing in the same way as before one gets A1  B1  A2  C  1,
2x 4  3 x 2  x
x 1
x
1
and B2  0. Hence 5



.
2
4
3
2
2
2
x  x  2x  2x  x  1 x  1 x  1
x 1


Integrating Partial Fraction Decompositions
After a general partial fraction decomposition one has to deal
with integrals of the following types. There are four cases.
Two first cases are easy.
A
A
1. 
dx  ln ax  b  K
ax  b
a
2.
1 l
A
  ax  b 
A  ax  b 
Here K is the constant of
integration.
l
dx 
a
1 l
 K , l  1.
In the remaining cases we have to compute integrals of the type:
Ax  B
3.  2
dx
ax  bx  c
and
4.

Ax  B
 ax
2
 bx  c

l
dx, l  1
We will discuss the integration of these cases based on
examples. Normally, after some transformations they result
in integrals which are either logarithms or tan-1.
Examples
Example 1
Compute
3
 x 3  1dx.
Observe x 3  1  ( x  1)( x 2  x  1). Hence
3
A
Bx  C


for some numbers A, B and C.
x3  1 x  1 x2  x  1
To compute these numbers A, B and C we get
3
A( x 2  x  1)
(Bx  C )( x  1)


x 3  1 ( x  1)( x 2  x  1) ( x 2  x  1)( x  1)

3
( A  B )x  ( A  B  C )x  A  C

x3  1
x3  1
2
Hence
3
1
x2


x3  1 x  1 x 2  x  1
AB 0

 A 1


  A  B  C  0   B  1.

C  2
A C  3


Examples
Example 1 (cont’d)
Compute
3
 x 3  1dx.
By the previous computations we now have
3
1
x2
dx

dx

 x3  1  x  1  x 2  x  1dx
 ln x  1 
1
2x  1
3
1
dx

dx
2  x2  x  1
2  x2  x  1

Substitute u=x2+x+1
in the first remaining
integral and rewrite
the last integral.

1
3
1
2
 ln x  1  ln x  x  1  
dx
2
2
2  x  1/ 2   3 / 4
1
 2x  1
 ln x  1  ln x 2  x  1  3 arctan 
K
2
3




This expression is the required substitution to finish the computation.
Examples
Example 2
x3  x  2
Compute 
dx.
2
x 1
We can simplify the function to be integrated by performing polynomial
division first. This needs to be done whenever possible. We get:
x3  x  2
2

x

x2  1
x2  1
Partial fraction decomposition for the remaining rational expression leads to
x3  x  2
2
1
1

x


x


x2  1
x2  1
x 1 x 1
Now we can integrate
x 3  1 2
1
1 

dx

x


 x2  1
  x  1 x  1 dx
x2
x2
x 1

 ln | x  1|  ln | x  1| K 
 ln
 K.
2
2
x 1