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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.
