Math123 Lecture 1Lecturer: Dr. Robert C. BusbyOffice: Korman 276Phone : 215-895-1957Email: rbusby@mcs.drexel.eduCourse Web Site: http://www.mcs.drexel.edu/classes/Calculus/MATH123_Spring02/

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Integration of Rational Functions using Partial Fractions

Example: We wish to evaluate the integral

5 10( 4)( 1)

x dxx x

−∫ − +

Integration of Rational Functions using Partial Fractions

Example: We wish to evaluate the integral

5 10( 4)( 1)

x dxx x

−∫ − +

As a first step, we need to break up the integrand into managablepieces. We will attempt to break the integrand into two pieces, each containing only one of the factors (x − 4)(x + 1) in its denominator. That is we want to write

5 10( 4)( 1) ( 4) ( 1)

x A Bx x x x

− = +− + − +

for some numbers A and B. If the identity is true, then by putting the terms on the right in a form where we have a common denominator, we get:

5 10( 4)( 1) ( 4) ( 1)

x A Bx x x x

− = +− + − +

for some numbers A and B. If the identity is true, then by putting the terms on the right in a form where we have a common denominator, we get:

5 10 ( 1) ( 4)( 4)( 1) ( 4)( 1)

x A x B xx x x x

− + + −=− + − +

5 10( 4)( 1) ( 4) ( 1)

x A Bx x x x

− = +− + − +

for some numbers A and B. If the identity is true, then by putting the terms on the right in a form where we have a common denominator, we get:

5 10 ( 1) ( 4)( 4)( 1) ( 4)( 1)

x A x B xx x x x

− + + −=− + − +

It then follows that 5 10 ( 1) ( 4).x A x B x− = + + −

This is not an equation to solve for x. It is an identity, meaning that it must be true for all possible values of x.

Method 1 for finding A and B. Collect like powers of xtogether on the right side of the equation

To get

It follows that 5 10 ( ) ( 4 )x A B x A B− = + + −( ) 5 and ( 4 ) 10A B A B+ = − =−

5 10 ( 1) ( 4).x A x B x− = + + −

Method 1 for finding A and B. Collect like powers of xtogether on the right side of the equation

To get

It follows that 5 10 ( ) ( 4 )x A B x A B− = + + −( ) 5 and ( 4 ) 10A B A B+ = − =−

5 10 ( 1) ( 4).x A x B x− = + + −

We can solve these equations simultaneously. The first can be written and substituted into the second to get5A B= −

(5 ) 4 10 or 5 15B B B− − =− =

It follows that 5B = 15, or B = 3. Thus A = 2 and so

5 10 2 3( 4)( 1) ( 4) ( 1)

xx x x x

− = +− + − +

Then

5 10 2 3 2ln 4 3ln 1( 4)( 1) ( 4) ( 1)

x dx dx dx x x Cx x x x

− = + = − + + +∫ ∫ ∫− + − +

5 10 2 3( 4)( 1) ( 4) ( 1)

xx x x x

− = +− + − +

Then

5 10 2 3 2ln 4 3ln 1( 4)( 1) ( 4) ( 1)

x dx dx dx x x Cx x x x

− = + = − + + +∫ ∫ ∫− + − +

Method 2 for finding A and B. substitute values of x into the identity

Any values will do, but the ones that produce the simplest results are x = 4 and x = –1. When substituted in, they produce the results and

Thus A = 2 and B = 3, as before.

5 10 ( 1) ( 4).x A x B x− = + + −

10 (5)A= 15 ( 5).B− = −

5 10 2 3( 4)( 1) ( 4) ( 1)

xx x x x

− = +− + − +

The terms on the right side of the equation

are called ‘partial fractions’.

5 10 2 3( 4)( 1) ( 4) ( 1)

xx x x x

− = +− + − +

The terms on the right side of the equation

are called ‘partial fractions’. Any rational function

( )( )

p xq x

with p(x) having smaller degrtee than q(x), can be written as the sum of partial fractions, and then integrated. We must first explore what partial fractions to use.

Types of partial fractions

Factoring a polynomial means breaking it up as the product of the smallest possible terms. Such terms may be ‘linear’, that isof the form

(ax + b)

or ‘quadratic’, that is of the form 2( )ax bx c+ +

Theorem: Every polynomial with real coefficients can be uniquely factored into linear factors and irreducible quadratic factors, that is quadratic factors that cannot be factored further without the use of complex numbers.

Types of partial fractions

Factoring a polynomial means breaking it up as the product of the smallest possible terms. Such terms may be ‘linear’, that isof the form

(ax + b)

or ‘quadratic’, that is of the form 2( )ax bx c+ +

( )max b+

Linear and quadratic factors can occur in a polynomial more that once. If a linear factor (ax + b) occurs m times then we will combine these factors together to produce

( )max b+

Linear and quadratic factors can occur in a polynomial more that once. If a linear factor (ax + b) occurs m times then we will combine these factors together to produce

If the irreducible quadratic factor occurs mtimes, then these factors are combined to produce

2( )ax bx c+ +

2( )max bx c+ +

( )max b+

Linear and quadratic factors can occur in a polynomial more that once. If a linear factor (ax + b) occurs m times then we will combine these factors together to produce

If the irreducible quadratic factor occurs mtimes, then these factors are combined to produce

2( )ax bx c+ +

2( )max bx c+ +Then we can restate the previous theorem as follows:

Theorem: Every polynomial with real coefficients can be uniquely factored into distinct factors of the form

and

with a, b, and c real.

( )max b+ 2( )max bx c+ +

Suppose that we wish to integrate a rational function

whose denominator has been factored as indicated in the previous theorem. It can be shown that the rational function canbe written as the sum of ‘partial fractions’ according to the following rules.

( )( )( )

p xr xq x

=

Suppose that we wish to integrate a rational function

whose denominator has been factored as indicated in the previous theorem. It can be shown that the rational function canbe written as the sum of ‘partial fractions’ according to the following rules.

Linear Factor Rule. For each factor of q having the form

the partial fraction decomposition contains a sum of the form

( )( )( )

p xr xq x

=

( )max b+

1 22( ) ( )

mA A Amax b ax b ax b

+ + ++ + +

L

Quadratic Factor Rule. For each factor of q having the form

The partial fraction decomposition contains a sum of the form

2( )max bx c+ +

1 1 2 22 2 2 2( ) ( ) ( )

x B m mA A x B A x Bmax bx c ax bx c ax bx c

+ + ++ + ++ + + + + +

L

If the denominators have quadratic factors, we use the following rule.

Example. Write out the form of the partial fraction expansion of the following rational functions, but do not solve for the unknown constants.

(a)

(b)

(c)

3 5( )2 2 2( 5)( 2)

x xr xx x

+ −=+ +

2 2 1( )2 2( 3)( 2) (3 1)

x xr xx x x x

+ +=+ − + +

4 32 5( )3 2 2( 6) ( 3) ( 5)

x xr xx x x

− +=+ − +

3 52 2 2 2 2 2 2( 5)( 2) ( 5) ( 2) ( 2)

x x Ax B Cx D Ex Fx x x x x

+ − + + += + ++ + + + +

Solution:

3 52 2 2 2 2 2 2( 5)( 2) ( 5) ( 2) ( 2)

x x Ax B Cx D Ex Fx x x x x

+ − + + += + ++ + + + +

Solution:

2 2 12 2 2 2( 3) ( 2)( 3)( 2) (3 1) ( 2) (3 1)

x x A B C Dx Ex xx x x x x x x

+ + += + + ++ −+ − + + − + +

3 52 2 2 2 2 2 2( 5)( 2) ( 5) ( 2) ( 2)

x x Ax B Cx D Ex Fx x x x x

+ − + + += + ++ + + + +

Solution:

2 2 12 2 2 2( 3) ( 2)( 3)( 2) (3 1) ( 2) (3 1)

x x A B C Dx Ex xx x x x x x x

+ + += + + ++ −+ − + + − + +

4 32 53 2 2( 6) ( 3) ( 5)

2 3 2 2( 6) ( 3) ( 5)( 6) ( 6) ( 3) ( 5)

x xx x x

A B C D E F Gx x xx x x x

− + =+ − +

+ + + + + ++ − ++ + − +

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

7( )( 3)( 5)

xr xx x

+=− +

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

7( )( 3)( 5)

xr xx x

+=− +

Solution. By the linear factor rule, the partial fractions decomposition must contain factors of the form

and ( 3) ( 5)

A Bx x− +

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

7( )( 3)( 5)

xr xx x

+=− +

Solution. By the linear factor rule, the partial fractions decomposition must contain factors of the form

and ( 3) ( 5)

A Bx x− +

Thus 7( 3)( 5) ( 3) ( 5)

x A Bx x x x

+ = +− + − +

We need to find the unknown coefficients A and B. To do this, put the right side over a common denominator, and equate numerators. We get

7 ( 5) ( 3)( 3)( 5) ( 3) ( 5) ( 3)( 5)

x A B A x B xx x x x x x

+ + + −= + =− + − + − +

so 7 ( 5) ( 3)x A x B x+ = + + −

We can then find A and B by substituting two convenient numbers into this identity. Any two will do, but the simplest results come from using 3 and –5. They are

7 ( 5) ( 3)( 3)( 5) ( 3) ( 5) ( 3)( 5)

x A B A x B xx x x x x x

+ + + −= + =− + − + − +

so 7 ( 5) ( 3)x A x B x+ = + + −

We can then find A and B by substituting two convenient numbers into this identity. Any two will do, but the simplest results come from using 3 and –5. They are

10 (8)A= (coming from x = 3) and (coming from x = – 5) 2 ( 8)B= −

Thus A = 5/4 and B = – ¼, therefore

7 5 1 1 1( 3)( 5) 4( 3) 4( 5)

xx x x x

+ = −− + − +

7 ( 5) ( 3)( 3)( 5) ( 3) ( 5) ( 3)( 5)

x A B A x B xx x x x x x

+ + + −= + =− + − + − +

so 7 ( 5) ( 3)x A x B x+ = + + −

We can then find A and B by substituting two convenient numbers into this identity. Any two will do, but the simplest results come from using 3 and –5. They are

10 (8)A= (coming from x = 3) and (coming from x = – 5) 2 ( 8)B= −

Thus A = 5/4 and B = – ¼, therefore

7 5 1 1 1( 3)( 5) 4( 3) 4( 5)

xx x x x

+ = −− + − +

7 1 55 1so ln 3 ln 5( 3)( 5) ( 3) 4 ( 5) 44 4

x dx dxdx x x Cx x x x

+ = − = − − + +∫∫ ∫− + − +

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

23 10( )2( 4)( 1)

x xr xx x

− −=− +

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

23 10( )2( 4)( 1)

x xr xx x

− −=− +

Solution. By the two factor rules, the partial fractions decomposition must contain factors of the form

and 2( 4) ( 1)

A Bx Cx x

+− +

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

23 10( )2( 4)( 1)

x xr xx x

− −=− +

Solution. By the two factor rules, the partial fractions decomposition must contain factors of the form

and 2( 4) ( 1)

A Bx Cx x

+− +

Thus23 10 +

2 2( 4)( 4)( 1) ( 1)

x x A Bx Cxx x x

− − +=−− + +

2 23 10 ( 1) ( )( 4)+2 2 2( 4)( 4)( 1) ( 1) ( 4)( 1)

x x A Bx C A x Bx C xxx x x x x

− − + + + + −= =−− + + − +

so 2 23 10 ( 1) ( )( 4)x x A x Bx C x− − = + + + −

2 23 10 ( 1) ( )( 4)+2 2 2( 4)( 4)( 1) ( 1) ( 4)( 1)

x x A Bx C A x Bx C xxx x x x x

− − + + + + −= =−− + + − +

so 2 23 10 ( 1) ( )( 4)x x A x Bx C x− − = + + + −

There are three unknowns, so we choose three values of x. We pick x = 4, x = 0, and x = 1. These yield respectively

34 (17)A= 10 4A C− = − 8 2 3( )A B C− = − +

2 23 10 ( 1) ( )( 4)+2 2 2( 4)( 4)( 1) ( 1) ( 4)( 1)

x x A Bx C A x Bx C xxx x x x x

− − + + + + −= =−− + + − +

so 2 23 10 ( 1) ( )( 4)x x A x Bx C x− − = + + + −

There are three unknowns, so we choose three values of x. We pick x = 4, x = 0, and x = 1. These yield respectively

34 (17)A= 10 4A C− = − 8 2 3( )A B C− = − +

Thus A = 2; 4C = 12, so C = 3; 3B = 2A – 3C + 8 = 3, so B = 1.

23 10 2 3+2 2( 4)( 4)( 1) ( 1)

x x xxx x x

− − +=−− + +

223 10 1+ 32 2 2( 4)( 4)( 1) ( 1) ( 1)

x x dx xdx dx dxxx x x x

− − = + ∫∫ ∫ ∫−− + + +

223 10 1+ 32 2 2( 4)( 4)( 1) ( 1) ( 1)

x x dx xdx dx dxxx x x x

− − = + ∫∫ ∫ ∫−− + + +

1 2 12ln 4 + ln 1 3tan2

x x x C−= − + + +

Thus we have the following procedure.

1. Start with a proper fraction r(x) (order of numerator less than order of denominator) with denominator of order k and factored.

2. Set up the corresponding partial fraction expansion on the right, with k unknowns.

3. Put the right side over a common denominator, and equate numerators. Choose and plug in k values of x. Pick values that make one factor vanish, and if needed pick the other values small.

4. Solve for the unknown constants.

5. Plug the constants into the partial fraction expansion and integrate (by substitution).

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

22 7( )2( 2) ( 1)

x xr xx x

+=+ −

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

22 7( )2( 2) ( 1)

x xr xx x

+=+ −

Solution. By the linear factor rule, the partial fractions decomposition must contain factors of the form

+ and 2( 2) ( 1)( 2)

A B Cx xx+ −+

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

22 7( )2( 2) ( 1)

x xr xx x

+=+ −

Solution. By the linear factor rule, the partial fractions decomposition must contain factors of the form

+ and 2( 2) ( 1)( 2)

A B Cx xx+ −+

Thus22 7 + +2 2( 2) ( 1)( 2) ( 1) ( 2)

x x A B Cx xx x x

+ =+ −+ − +

2( 2)( 1) ( 1) ( 2)+ +2 2( 2) ( 1)( 2) ( 2) ( 1)

A B C A x x B x C xx xx x x

+ − + − + +=+ −+ + −

so 2 22 7 ( 2)( 1) ( 1) ( 2)x x A x x B x C x+ = + − + − + +

We need three values of x to get three equations for A, B, and C.We choose x = 1 and x = – 2, as usual, then choose a convenient small x for the third. We choose x = 0.

2( 2)( 1) ( 1) ( 2)+ +2 2( 2) ( 1)( 2) ( 2) ( 1)

A B C A x x B x C xx xx x x

+ − + − + +=+ −+ + −

so 2 22 7 ( 2)( 1) ( 1) ( 2)x x A x x B x C x+ = + − + − + +

We need three values of x to get three equations for A, B, and C.We choose x = 1 and x = – 2, as usual, then choose a convenient small x for the third. We choose x = 0. The results are

6 ( 3)B− = −9 (9)C= 0 ( 2) 4A B C= − − +

Thus C = 1; B = 2; A = (B – 4C )/ – 2 = 1, so

22 7 1 2 1+ +2 2( 2) ( 1)( 2) ( 1) ( 2)

x xx xx x x

+ =+ −+ − +

22 7 + 2 +2 2( 2) ( 1)( 2) ( 1) ( 2)

x x dx dx dxdxx xx x x

+ =∫ ∫ ∫ ∫+ −+ − +

22 7 + 2 +2 2( 2) ( 1)( 2) ( 1) ( 2)

x x dx dx dxdxx xx x x

+ =∫ ∫ ∫ ∫+ −+ − +

2ln 2 +ln 1( 2)

x x Cx

= + − − ++

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

3 2 2( )2 2( 1)( 2)

x x xr xx x

+ + +=+ +

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

3 2 2( )2 2( 1)( 2)

x x xr xx x

+ + +=+ +

Solution. By the linear factor rule, the partial fractions decomposition must contain factors of the form

+ and 2 2( 1) ( 2)

Ax B Cx Dx x

+ +

+ +

Example. Find the partial fraction decomposition of

and integrate the resulting expression.

3 2 2( )2 2( 1)( 2)

x x xr xx x

+ + +=+ +

Solution. By the linear factor rule, the partial fractions decomposition must contain factors of the form

+ and 2 2( 1) ( 2)

Ax B Cx Dx x

+ +

+ +Thus

3 2 2 +2 2 2 2( 1)( 2) ( 1) ( 2)

x x x Ax B Cx Dx x x x

+ + + + +=+ + + +

3 2 2 22 ( )( 2) ( )( 1)+2 2 2 2 2 2( 1)( 2) ( 1) ( 2) ( 1)( 2)

x x x Ax B Cx D Ax B x Cx D xx x x x x x

+ + + + + + + + + += =+ + + + + +

3 2 2 22 ( )( 2) ( )( 1)x x x Ax B x Cx D x+ + + = + + + + +

so

3 2 2 22 ( )( 2) ( )( 1)+2 2 2 2 2 2( 1)( 2) ( 1) ( 2) ( 1)( 2)

x x x Ax B Cx D Ax B x Cx D xx x x x x x

+ + + + + + + + + += =+ + + + + +

3 2 2 22 ( )( 2) ( )( 1)x x x Ax B x Cx D x+ + + = + + + + +

so

Here we cannot make any factors vanish by a choice of x, so we simply need to pick 4 small values of x. We choose −2,−1, 0, and 1. The resulting equations are

1 3( ) 2( )A B C D= − + + − +

2 2B D= +

For x = −1

For x = 0

5 3( ) 2( )A B C D= + + + For x = 1

4 6( 2 ) 5( 2 )A B C D− = − + + − + For x = −2

(2) 3 3 2 2 1A B C D− + − + =(3) 2 2B D+ =(4) 3 3 2 2 5A B C D+ + + =

(1) 12 6 10 5 4A B C D− + − + =−

(2) 3 3 2 2 1A B C D− + − + =(3) 2 2B D+ =(4) 3 3 2 2 5A B C D+ + + =

(1) 12 6 10 5 4A B C D− + − + =−

(2)+(4) 6 4 6B D+ =(3) 6 3 6B D+ =

D = 0

(1) 12 10 10A C− − =−(2) 3 2 2A C− − =− (2) 12 8 8A C− − =−

B = 1

C = 1

A = 0

Thus 3 2 2 1 +2 2 2 2( 1)( 2) ( 1) ( 2)

x x x xx x x x

+ + + =+ + + +

so3 2 2 12 2 2 2( 1)( 2) ( 1) ( 2)

x x x xdx dx dxx x x x

+ + + = +∫ ∫ ∫+ + + +

Thus 3 2 2 1 +2 2 2 2( 1)( 2) ( 1) ( 2)

x x x xx x x x

+ + + =+ + + +

so3 2 2 12 2 2 2( 1)( 2) ( 1) ( 2)

x x x xdx dx dxx x x x

+ + + = +∫ ∫ ∫+ + + +

11tan ( )2

duxu

−= + ∫

11 2tan ( ) ln( 1)2

x x C−= + + +