Partial Fraction Decompositions

Partial Fraction Decompositions

In this and 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.

Partial Fraction Decompositions

We assume the rational functions are reduced.

In this and 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.

Partial Fraction Decompositions

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.

In this and 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.

Partial Fraction DecompositionsExample:

x3 x2 + x + 1

=

By long division

Partial Fraction DecompositionsExample:

x3 x2 + x + 1

=1

x – 1 +

By long division

x2 + x + 1

Partial Fraction DecompositionsExample:

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

Partial Fraction DecompositionsExample:

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

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

Partial Fraction DecompositionsExample:

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

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.

Partial Fraction DecompositionsExample:

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

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 irrducible.

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 irrducible.

Partial Fraction Decomposition Theorem:

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 irrducible.

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 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 irrducible.

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 Decompositions

Example: a. For P(x)(x + 2)(x – 3)

Partial Fraction Decompositions

Example: a. For P(x)(x + 2)(x – 3)

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

the denominator

Partial Fraction Decompositions

Example: a. For P(x)(x + 2)(x – 3)

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

(x + 2)(x – 3) =A

(x + 2) +B

(x – 3 )

the denominator

Partial Fraction Decompositions

Example: a. For P(x)(x + 2)(x – 3)

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

Partial Fraction Decompositions

Example: a. For P(x)(x + 2)(x – 3)

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

, the denominator may be

Partial Fraction Decompositions

Example: a. For P(x)(x + 2)(x – 3)

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

, the denominator may be

+ C(x – 3)2

+ D(x – 3)3

Partial Fraction Decompositions

c. For P(x)(x + 2)(x2 + 3) linear factors (x + 2) and a 2nd degree irreducible

factor (x2 + 3),

the denominator has one

Partial Fraction Decompositions

c. For P(x)(x + 2)(x2 + 3) linear factors (x + 2) and a 2nd degree irreducible

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

(x + 2)(x2 + 3)

=A

(x + 2) +Bx + C

(x2 + 3)

the denominator has one

Partial Fraction Decompositions

c. For P(x)(x + 2)(x2 + 3) linear factors (x + 2) and a 2nd degree irreducible

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

(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,

the denominator has one

, the denominator may be

Partial Fraction Decompositions

c. For P(x)(x + 2)(x2 + 3) linear factors (x + 2) and a 2nd degree irreducible

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

(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

the denominator has one

, the denominator may be

Partial Fraction Decompositions

c. For P(x)(x + 2)(x2 + 3) linear factors (x + 2) and a 2nd degree irreducible

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

(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

the denominator has one

, the denominator may be

+ 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.

Partial Fraction Decompositions

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)

Partial Fraction Decompositions

Example: Find the partial fraction

decomposition of

We know that

(x + 2)(x – 3) =A

(x + 2) +B

(x – 3 )

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)

2x +3

Partial Fraction Decompositions

Example: Find the partial fraction

decomposition of

We know that

(x + 2)(x – 3) =A

(x + 2) +B

(x – 3 )

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)

2x +3

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

Partial Fraction Decompositions

Example: Find the partial fraction

decomposition of

We know that

(x + 2)(x – 3) =A

(x + 2) +B

(x – 3 )

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)

2x +3

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

We have 2x + 3 = A(x – 3) + B(x + 2)

Partial Fraction Decompositions

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

Partial Fraction Decompositions

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,

Partial Fraction Decompositions

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,

We have 9 = 0 + 5B

Partial Fraction Decompositions

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,

or B = 9/5

We have 9 = 0 + 5B

Partial Fraction Decompositions

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,

Evaluate x = -2,

or B = 9/5

2x + 3 = A(x – 3) + B(x + 2)

We have 9 = 0 + 5B

Partial Fraction Decompositions

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,

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

Partial Fraction Decompositions

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,

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

Partial Fraction Decompositions

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,

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.

Partial Fraction Decompositions

(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3

We know that

1

Example: Find the partial fraction

decomposition of 1

(x – 2)(x – 3)3.

Partial Fraction Decompositions

(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)

Example: Find the partial fraction

decomposition of 1

(x – 2)(x – 3)3.

Partial Fraction Decompositions

(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3

Example: Find the partial fraction

decomposition of 1

(x – 2)(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)

Evaluate at x = 3,

Partial Fraction Decompositions

(x – 2)(x – 3)3

= A(x – 2)

+ B(x – 3)

+ C(x – 3)2

+ D(x – 3)3

Example: Find the partial fraction

decomposition of 1

(x – 2)(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)

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

Partial Fraction Decompositions

(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)

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

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

Example: Find the partial fraction

decomposition of 1

(x – 2)(x – 3)3.

Partial Fraction Decompositions

(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)

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

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

These are the only roots we can use to evaluate.

Example: Find the partial fraction

decomposition of 1

(x – 2)(x – 3)3.

Partial Fraction Decompositions

(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)

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

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

These are the only roots we can use to evaluate.

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

Example: Find the partial fraction

decomposition of 1

(x – 2)(x – 3)3.

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

Partial Fraction Decompositions

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

Partial Fraction Decompositions

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

Partial Fraction Decompositions

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 = -x3 + 3x2 – 3x + 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

Partial Fraction Decompositions

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

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

Partial Fraction Decompositions

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

x3 – 3x2 + 2x – 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

Partial Fraction Decompositions

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

x3 – 3x2 + 2x – 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

Partial Fraction Decompositions

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

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

x3 + …. = Bx3 + …..

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

Partial Fraction Decompositions

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

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

x3 + …. = Bx3 + …..

Hence B = 1,

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

Partial Fraction Decompositions

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

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

x3 + …. = Bx3 + …..

Hence B = 1, put this back into the equation, we have

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

Partial Fraction Decompositions

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

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

x3 + …. = Bx3 + …..

Hence B = 1, put this back into the equation, we have

x3 – 3x2 + 2x – 24 = 1 (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

Partial Fraction Decompositions

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

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

x3 + …. = Bx3 + …..

Hence B = 1, put this back into the equation, we have

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

Match the constant terms from both sides

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

Partial Fraction Decompositions

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

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

x3 + …. = Bx3 + …..

Hence B = 1, put this back into the equation, we have

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

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

Partial Fraction Decompositions

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

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

x3 + …. = Bx3 + …..

Hence B = 1, put this back into the equation, we have

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

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

Partial Fraction Decompositions

(x – 2)(x – 3)3

= -1(x – 2)

+ 1(x – 3)

+ -1(x – 3)2

+ 1(x – 3)3.

Therefore1

Partial Fraction Decompositions

(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

Partial Fraction Decompositions

(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

Partial Fraction Decompositions

(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

Partial Fraction Decompositions

(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)

Evaluate at x = -2,

1 – 2x

Partial Fraction Decompositions

(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)

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

1 – 2x

Partial Fraction Decompositions

(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)

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)

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 1

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 1

Compare the square terms,

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 1

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

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 1

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

Hence B = -1

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 1

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

Hence B = -1 and that 1 – 2x = -2x + Cx + 2C + 1

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 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

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 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

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 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 C = 0.

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

1 – 2x = x2 + 1 + Bx2 + 2Bx + Cx + 2C

1 – 2x = Bx2 + x2 + 2Bx + Cx + 2C + 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 – 2x (x + 2)(x2 + 1)

=1

(x + 2) +-x

(x2 + 1).

Therefore

1 = 2C + 1 C = 0.