X19 partial fraction decompositions

$: X19 partial fraction decompositions$
Partial Fraction Decompositions


In this and the next sections we will show how to integrate rational functions, that is, functions of the form P(x)

Q(x) where P and Q are polynomials.


We assume the rational functions are reduced.




We assume the rational functions are reduced.

Furthermore we assume that deg Q > deg P, for if not, we can use long division to reduce the problem to integrating such a function.



Partial Fraction DecompositionsExample:

x3 x2 + x + 1 = 1 x – 1 +

By long division

x2 + x + 1


x3 x2 + x + 1 = 1 x – 1 +

By long division

So finding the integral of

x2 + x + 1 x3

x2 + x + 1 is reduced to

finding the integral of . 1

x2 + x + 1


x3 x2 + x + 1 = 1 x – 1 +

By long division


x2 + x + 1 x3



x2 + x + 1

To integrate P/Q, we break it down as the sum of smaller rational formulas.


x3 x2 + x + 1 = 1 x – 1 +

By long division


x2 + x + 1 x3



x2 + x + 1

To integrate P/Q, we break it down as the sum of smaller rational formulas. This is called the partial fraction decomposition of P/Q.


x3 x2 + x + 1 = 1 x – 1 +

By long division


x2 + x + 1 x3



x2 + x + 1

To integrate P/Q, we break it down as the sum of smaller rational formulas. This is called the partial fraction decomposition of P/Q. This decomposition is unique.

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors.

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)a. x6 – x2

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)a. x6 – x2 = x2(x4 – 1)

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)a. x6 – x2 = x2(x4 – 1) = x2(x2 – 1)(x2 + 1)

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)a. x6 – x2 = x2(x4 – 1) = x2(x2 – 1)(x2 + 1) = x2(x – 1)(x + 1)(x2 + 1)

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)a. x6 – x2 = x2(x4 – 1) = x2(x2 – 1)(x2 + 1) = x2(x – 1)(x + 1)(x2 + 1)b. x6 – 1

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)a. x6 – x2 = x2(x4 – 1) = x2(x2 – 1)(x2 + 1) = x2(x – 1)(x + 1)(x2 + 1)b. x6 – 1 = (x3 – 1)(x3 + 1)

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)a. x6 – x2 = x2(x4 – 1) = x2(x2 – 1)(x2 + 1) = x2(x – 1)(x + 1)(x2 + 1)b. x6 – 1 = (x3 – 1)(x3 + 1) = (x – 1)(x2 + x + 1)(x + 1)(x2 – x + 1)

Partial Fraction DecompositionsFundamental Theorem of Algebra: Given any real (coefficient) polynomial Q(x), it can be written as the product of1st degree and irreducible 2nd degree factors. Example: Factor completely (over real numbers)a. x6 – x2 = x2(x4 – 1) = x2(x2 – 1)(x2 + 1) = x2(x – 1)(x + 1)(x2 + 1)b. x6 – 1 = (x3 – 1)(x3 + 1) = (x – 1)(x2 + x + 1)(x + 1)(x2 – x + 1)c. x2 – 2 = (x – 2)(x + 2)

Partial Fraction DecompositionsReminder: To check if ax2 + bx + c is reducible,check b2 – 4ac.

Partial Fraction DecompositionsReminder: To check if ax2 + bx + c is reducible,check b2 – 4ac. If b2 – 4ac > 0, it is reducible.

Partial Fraction DecompositionsReminder: To check if ax2 + bx + c is reducible,check b2 – 4ac. If b2 – 4ac > 0, it is reducible.If b2 – 4ac < 0, it is irreducible.


Partial Fraction Decomposition Theorem:


Partial Fraction Decomposition Theorem:Given P/Q where deg P < deg Q, then P/Q = F1 + F2 + .. + Fn where

Fi = or (ax + b)k (ax2 + bx + c)k

(Ai and Bi are numbers)Aix + BiAi


Partial Fraction Decomposition Theorem:Given P/Q where deg P < deg Q, then P/Q = F1 + F2 + .. + Fn where

Fi = or Ai

(ax + b)k Aix + Bi

(ax2 + bx + c)k and that (ax + b)k or (ax2 + bx + c)k are

factors in the factorization of Q(x).

(Ai and Bi are numbers)

Partial Fraction DecompositionsExample: a. For P(x)

(x + 2)(x – 3)


(x + 2)(x – 3) has two linear factor (x + 2), (x – 3),

the denominator


(x + 2)(x – 3) has two linear factor (x + 2), (x – 3), therefore

P(x)(x + 2)(x – 3) = A

(x + 2) + B(x – 3 )

the denominator



P(x)(x + 2)(x – 3) = A

(x + 2) + B(x – 3 )

b. For P(x)(x + 2)(x – 3)3 viewed having two factors (x + 2), (x – 3)3,

the denominator

, the denominator may be



P(x)(x + 2)(x – 3) = A

(x + 2) + B(x – 3 )

b. For P(x)(x + 2)(x – 3)3 viewed having two factors (x + 2), (x – 3)3, hence

the denominators in the decomposition are(x + 2), (x – 3), (x – 3)2, (x – 3)3 and

the denominator




P(x)(x + 2)(x – 3) = A

(x + 2) + B(x – 3 )

b. For P(x)(x + 2)(x – 3)3 viewed having two factors (x + 2), (x – 3)3, hence

the denominators in the decomposition are(x + 2), (x – 3), (x – 3)2, (x – 3)3 and

P(x)(x + 2)(x – 3)3

= A(x + 2)

+ B(x – 3)

the denominator


+ C(x – 3)2

+ D(x – 3)3

Partial Fraction Decompositionsc. For P(x)

(x + 2)(x2 + 3) linear factor (x + 2) and a 2nd degree irreducible

factor (x2 + 3),

the denominator has one



factor (x2 + 3), hence P(x)

(x + 2)(x2 + 3)

= A(x + 2) + Bx + C

(x2 + 3)





(x + 2)(x2 + 3)

= A(x + 2) + Bx + C

(x2 + 3)

d. For P(x)(x + 2)2(x2 + 3)2

viewed having two factors (x + 2)2, (x2 + 3)2,






(x + 2)(x2 + 3)

= A(x + 2) + Bx + C

(x2 + 3)

d. For P(x)(x + 2)2(x2 + 3)2

viewed having two factors (x + 2)2, (x2 + 3)2, hencethe denominators in the decomposition are(x + 2), (x + 2)2 , (x2 + 3), (x2 + 3)2, hence






(x + 2)(x2 + 3)

= A(x + 2) + Bx + C

(x2 + 3)

d. For P(x)(x + 2)2(x2 + 3)2

viewed having two factors (x + 2)2, (x2 + 3)2, hencethe denominators in the decomposition are(x + 2), (x + 2)2 , (x2 + 3), (x2 + 3)2, hence

= A(x + 2)

+ B(x + 2)2



+ Cx + D(x2 + 3)

+ Ex + F(x2 + 3)2

P(x)(x + 2)2(x2 + 3)2

Partial Fraction DecompositionsTo find the exact decomposition, we use two methods:

Partial Fraction DecompositionsTo find the exact decomposition, we use two methods:1. by evaluation with the roots of the LCD

Partial Fraction DecompositionsTo find the exact decomposition, we use two methods:1. by evaluation with the roots of the LCD2. use the answers from 1, expand and match coefficients.


Example: Find the partial fraction

decomposition of

To find the exact decomposition, we use two methods:1. by evaluation with the roots of the LCD2. use the answers from 1, expand and match coefficients.

2x +3(x + 2)(x – 3)



decomposition of

We know that

(x + 2)(x – 3) = A(x + 2) + B

(x – 3 )


2x +3(x + 2)(x – 3)

2x +3



decomposition of

We know that

(x + 2)(x – 3) = A(x + 2) + B

(x – 3 )


2x +3(x + 2)(x – 3)

2x +3

Clear the denominator, multiply it by (x + 2)(x – 3)



decomposition of

We know that

(x + 2)(x – 3) = A(x + 2) + B

(x – 3 )


2x +3(x + 2)(x – 3)

2x +3

Clear the denominator, multiply it by (x + 2)(x – 3)We have 2x + 3 = A(x – 3) + B(x + 2)

Partial Fraction DecompositionsThe factors (x – 3), (x + 2) have roots at x = 3, x = -2


2x + 3 = A(x – 3) + B(x + 2)

The factors (x – 3), (x + 2) have roots at x = 3, x = -2

Evaluate at x = 3,


2x + 3 = A(x – 3) + B(x + 2)


Evaluate at x = 3,

We have 9 = 0 + 5B


2x + 3 = A(x – 3) + B(x + 2)


Evaluate at x = 3,

or B = 9/5

We have 9 = 0 + 5B


2x + 3 = A(x – 3) + B(x + 2)


Evaluate at x = 3,

Evaluate x = -2,

or B = 9/5

2x + 3 = A(x – 3) + B(x + 2)

We have 9 = 0 + 5B


2x + 3 = A(x – 3) + B(x + 2)


Evaluate at x = 3,

Evaluate x = -2,

or B = 9/5

2x + 3 = A(x – 3) + B(x + 2)

We have -1 = -5A + 0

We have 9 = 0 + 5B


2x + 3 = A(x – 3) + B(x + 2)


Evaluate at x = 3,

Evaluate x = -2,

or B = 9/5

2x + 3 = A(x – 3) + B(x + 2)

or A = 1/5

We have 9 = 0 + 5B

We have -1 = -5A + 0


2x + 3 = A(x – 3) + B(x + 2)


Evaluate at x = 3,

Evaluate x = -2,

or B = 9/5

2x + 3 = A(x – 3) + B(x + 2)

or A = 1/5

Hence (x + 2)(x – 3) = 1/5(x + 2) + 9/5

(x – 3 ) . 2x +3

We have 9 = 0 + 5B

We have -1 = -5A + 0

Partial Fraction DecompositionsExample: Find the partial fraction

decomposition of 1 (x – 2)(x – 3)3.


(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3

We know that1




(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3

We know that

Clear the denominator, We have

1

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




(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3



We know that


1

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

Evaluate at x = 3,


(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3



We know that


1

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

Evaluate at x = 3, we have 1 = D.


(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3

We know that


1

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


Evaluate at x = 2, we have -1= A.




(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3

We know that


1

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



These are the only roots we can use to evaluate.




(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3

We know that


1

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



These are the only roots we can use to evaluate.

For B and C, we use the “method of coefficient–matching”.



Partial Fraction Decompositions1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve


Put A = -1 and D = 1 into the equation and expand.1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2) Put A = -1 and D = 1 into the equation and expand.

1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve



1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve



move all the explicit terms to one side

1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2



x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3) move all the explicit terms to one side

1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2)

match the highest degree term from both sides

Put A = -1 and D = 1 into the equation and expand.


1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3)


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2)



x3 + …. = Bx3 + …..


1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3)


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2)



x3 + …. = Bx3 + ….. Hence B = 1,


1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3)


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2)



x3 + …. = Bx3 + ….. Hence B = 1, put this back into the equation, we have


1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3)


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2)




x3 – 3x2 + 2x – 24 = 1 (x – 2) (x – 3)2 + C(x – 2) (x – 3)


1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3)


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2)




x3 – 3x2 + 2x – 24 = 1 (x – 2) (x – 3)2 + C(x – 2) (x – 3)

Match the constant terms from both sides


1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3)


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2)




x3 – 3x2 + 2x – 24 = 1 (x – 2) (x – 3)2 + C(x – 2) (x – 3)

Match the constant terms from both sides…… – 24 = ….. – 18 + Cx2 – 5Cx + 6C


1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3)


1 = (-1)(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +(1)(x – 2)




x3 – 3x2 + 2x – 24 = 1 (x – 2) (x – 3)2 + C(x – 2) (x – 3)

Match the constant terms from both sides…… – 24 = ….. – 18 + Cx2 – 5Cx + 6C

We have -24 = -18 + 6C -1 = C


1 = A(x – 3)3 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) +D(x – 2) We’ve

1 = -x3 + 9x2 – 9x + 27 + B(x – 2) (x – 3)2 + C(x – 2) (x – 3) + x – 2

x3 – 9x2 + 26x – 24 =

B(x – 2) (x – 3)2 + C(x – 2) (x – 3)


(x – 2)(x – 3)3

= -1(x – 2)

+ 1(x – 3)

+ -1(x – 3)2

+ 1(x – 3)3.

Therefore1


(x – 2)(x – 3)3

= -1(x – 2)

+ 1(x – 3)

+ -1(x – 3)2

+ 1(x – 3)3.

Therefore1

(x + 2)(x2 + 1)

Example: Find the decomposition of 1 – 2x


(x – 2)(x – 3)3

= -1(x – 2)

+ 1(x – 3)

+ -1(x – 3)2

+ 1(x – 3)3.

Therefore1

(x + 2)(x2 + 1) We have

1 – 2x (x + 2)(x2 + 1)

= A(x + 2) + Bx + C

(x2 + 1)

Example: Find the decomposition of 1 – 2x


(x – 2)(x – 3)3

= -1(x – 2)

+ 1(x – 3)

+ -1(x – 3)2

+ 1(x – 3)3.

Therefore1

(x + 2)(x2 + 1) We have

1 – 2x (x + 2)(x2 + 1)

= A(x + 2) + Bx + C

(x2 + 1)

Example: Find the decomposition of

Clear the denominator, we've

1 – 2x =

A(x2 + 1)

+ (Bx + C)(x + 2)

1 – 2x


(x – 2)(x – 3)3

= -1(x – 2)

+ 1(x – 3)

+ -1(x – 3)2

+ 1(x – 3)3.

Therefore1

(x + 2)(x2 + 1) We have

1 – 2x (x + 2)(x2 + 1)

= A(x + 2) + Bx + C

(x2 + 1)



1 – 2x =

A(x2 + 1)

+ (Bx + C)(x + 2)

Evaluate at x = -2,

1 – 2x


(x – 2)(x – 3)3

= -1(x – 2)

+ 1(x – 3)

+ -1(x – 3)2

+ 1(x – 3)3.

Therefore1

(x + 2)(x2 + 1) We have

1 – 2x (x + 2)(x2 + 1)

= A(x + 2) + Bx + C

(x2 + 1)



1 – 2x =

A(x2 + 1)

+ (Bx + C)(x + 2)

Evaluate at x = -2, We have 5 = 5A or A = 1.

1 – 2x


(x – 2)(x – 3)3

= -1(x – 2)

+ 1(x – 3)

+ -1(x – 3)2

+ 1(x – 3)3.

Therefore1

(x + 2)(x2 + 1) We have

1 – 2x (x + 2)(x2 + 1)

= A(x + 2) + Bx + C

(x2 + 1)



1 – 2x =

A(x2 + 1)

+ (Bx + C)(x + 2)

Evaluate at x = -2, We have 5 = 5A or A = 1.

So we've 1 – 2x = x2 + 1 + (Bx + C)(x + 2)

1 – 2x

Partial Fraction DecompositionsExpand 1 – 2x = x2 + 1 + (Bx + C)(x + 2)


1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C


1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 1



Compare the square terms,



Compare the square terms, we've Bx2 + x2 = 0.



Compare the square terms, we've Bx2 + x2 = 0.Hence B = -1



Compare the square terms, we've Bx2 + x2 = 0.Hence B = -1 and that 1 – 2x = -2x + Cx + 2C + 1




Compare the constant terms





1 = 2C + 1





1 = 2C + 1 C = 0.





1 – 2x (x + 2)(x2 + 1)

= 1(x + 2) + -x

(x2 + 1).

Therefore

1 = 2C + 1 C = 0.


The fractional form 1 xK(1 – x)M(1 + x)M.

appears in the integration of the quotients of powers of sine and cosine.Example: Find the decomposition of

1 x(1 – x)(1 + x)


x(1 – x) (1 + x) = Ax + B

(1 – x) +

The decomposition has the form 1



1 x(1 – x)(1 + x)

C(1 + x)


x(1 – x) (1 + x) = Ax + B

(1 – x) +


1 = A(1 – x) (1 + x) + Bx(1 + x) + Cx(1 – x)



1 x(1 – x)(1 + x)

C(1 + x)

Clearing the denominator, we have


x(1 – x) (1 + x) = Ax + B

(1 – x) +


1 = A(1 – x) (1 + x) + Bx(1 + x) + Cx(1 – x)



1 x(1 – x)(1 + x)

C(1 + x)


Set x = 0, 1, –1, we obtain that A = 1, B = 1/2, C = –1/2 respectively,


x(1 – x) (1 + x) = Ax + B

(1 – x) +


1 = A(1 – x) (1 + x) + Bx(1 + x) + Cx(1 – x)



1 x(1 – x)(1 + x)

C(1 + x)


Set x = 0, 1, –1, we obtain that A = 1, B = 1/2, C = –1/2 respectively, hence

x(1 – x) (1 + x) = 1

x + 1/2(1 – x) –1 1/2

(1 + x)




1 x(1 – x)2(1 + x)2


x(1 – x)2(1 + x)2 =Ax + B

(1 – x) + D The decomposition has the form

1



1 x(1 – x)2(1 + x)2

C(1 – x)2 + (1 + x) +

E (1 + x)2


x(1 – x)2(1 + x)2 =Ax + B


1

1 = A(1 – x)2(1 + x)2 + Bx(1 – x)(1 + x)2 + Cx(1 + x)2 + Dx(1 – x)2(1 + x) + Ex(1 – x)2



1 x(1 – x)2(1 + x)2

C(1 – x)2 + (1 + x) +

E (1 + x)2

Clear the denominator, we have


x(1 – x)2(1 + x)2 =Ax + B


1

1 = A(1 – x)2(1 + x)2 + Bx(1 – x)(1 + x)2 + Cx(1 + x)2 + Dx(1 – x)2(1 + x) + Ex(1 – x)2



1 x(1 – x)2(1 + x)2

C(1 – x)2 + (1 + x) +

E (1 + x)2

Clear the denominator, we have

Set x = 0, 1, –1, we obtain A = 1, C = 1/4, E = –1/4.

Partial Fraction DecompositionsSet x = 0, 1, –1, we obtain A = 1, C = 1/4, E = –1/4. Hence1 = (1 – x)2(1 + x)2 + Bx(1 – x)(1 + x)2 + ¼ x(1 + x)2 + Dx(1 – x)2(1 + x) – ¼ x(1 – x)2


(1 – x)2(1 + x)2 + ¼ x(1 + x)2 – ¼ x(1 – x)2

Expand the known parts by grouping them first:

Set x = 0, 1, –1, we obtain A = 1, C = 1/4, E = –1/4. Hence1 = (1 – x)2(1 + x)2 + Bx(1 – x)(1 + x)2 + ¼ x(1 + x)2 + Dx(1 – x)2(1 + x) – ¼ x(1 – x)2


(1 – x)2(1 + x)2 + ¼ x(1 + x)2 – ¼ x(1 – x)2 = (1 – x2)2 + ¼ x[ (1 + x)2 – (1 – x)2]




(1 – x)2(1 + x)2 + ¼ x(1 + x)2 – ¼ x(1 – x)2 = (1 – x2)2 + ¼ x[ (1 + x)2 – (1 – x)2] = 1 – 2x2 + x4 + ¼ x[(1 + x) – (1 – x)] [(1 + x) + (1 – x)]




(1 – x)2(1 + x)2 + ¼ x(1 + x)2 – ¼ x(1 – x)2 = (1 – x2)2 + ¼ x[ (1 + x)2 – (1 – x)2] = 1 – 2x2 + x4 + ¼ x[(1 + x) – (1 – x)] [(1 + x) + (1 – x)] = 1 – 2x2 + x4 + ¼ x[2x][2]= 1 – x2 + x4







Hence,1 = 1 – x2 + x4 + Bx(1 – x)(1 + x)2 + Dx(1 – x)2(1 + x)





Hence,1 = 1 – x2 + x4 + Bx(1 – x)(1 + x)2 + Dx(1 – x)2(1 + x) or that x2 – x4 = Bx(1 – x)(1 + x)2 + Dx(1 – x)2(1 + x)





Hence,1 = 1 – x2 + x4 + Bx(1 – x)(1 + x)2 + Dx(1 – x)2(1 + x) or that x2 – x4 = Bx(1 – x)(1 + x)2 + Dx(1 – x)2(1 + x)Compare the linear terms, we have 0 = Bx + Dx so that D = –B

Partial Fraction DecompositionsHencex2 – x4 = Bx(1 – x)(1 + x)2 – Bx(1 – x)2(1 + x)


Compare the x4 terms, we have –x4 = –Bx4 – Bx4 so that B = 1/2, and D = –1/2.



x(1 – x)2(1 + x)2 =1 x + 1/2

(1 – x) + 1/2

Therefore1 1/4

(1 – x)2 (1 + x) – 1/4 (1 + x)2 –



x(1 – x)2(1 + x)2 =1 x + 1/2

(1 – x) + 1/2

Therefore1 1/4

(1 – x)2 (1 + x) – 1/4 (1 + x)2 –

In light of the last two examples, what do you think is the decomposition of the rational forms:

1 x(1 – x)M(1 + x)M

X19 partial fraction decompositions

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