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Part= ial Fraction Decompositions
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Part= ial Fraction Decompositions
In this and next sections we will show how to integrate rational functions, that is, functions = of the form  <= /span>        = ;       
P(x)
Q(x) <= /span>
where= P and Q are polynomials.
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Part= ial Fraction Decompositions
We a= ssume the rational functions are reduced.
In this and next sections we will show how to integrate rational functions, that is, functions = of the form  <= /span>        = ;       
P(x)
Q(x) <= /span>
where= P and Q are polynomials.
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Part= ial Fraction Decompositions
We a= ssume 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  <= /span>        = ;       
P(x)
Q(x) <= /span>
where= P and Q are polynomials.
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Part= ial Fraction Decompositions
Example:
x3 
x2 + x + 1
=3D
By long division
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Part= ial Fraction Decompositions
Example:
x3 
x2 + x + 1
=3D
1
x – 1 +
By long division
x2 + x + 1
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Part= ial Fraction Decompositions
Example:
x3 
x2 + x + 1
=3D
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
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Part= ial Fraction Decompositions
Example:
x3 
x2 + x + 1
=3D
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.
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Part= ial Fraction Decompositions
Example:
x3 
x2 + x + 1
=3D
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. 
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Part= ial Fraction Decompositions
Example:
x3 
x2 + x + 1
=3D
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.
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real number= s)
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real numbers)
a. x6 – x2 
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real numbers)
a. x6 – x2 =3D x2(x4 – 1)
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real numbers)
a. x6 – x2 =3D x2(x4 – 1)
        = ;       =3D x2(x2 – 1)(x2 + 1)
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real numbers)
a. x6 – x2 =3D x2(x4 – 1)
        = ;       =3D x2(x2 – 1)(x2 + 1)
        = ;       =3D x2(x – 1)(x + 1)(x2 + 1)
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real numbers)
a. x6 – x2 =3D x2(x4 – 1)
        = ;       =3D x2(x2 – 1)(x2 + 1)
        = ;       =3D x2(x – 1)(x + 1)(x2 + 1)
b. x6 – 1
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real numbers)
a. x6 – x2 =3D x2(x4 – 1)
        = ;       =3D x2(x2 – 1)(x2 + 1)
        = ;       =3D x2(x – 1)(x + 1)(x2 + 1)
b. x6 – 1 =3D (x3 – 1)(x3 + 1)
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real numbers)
a. x6 – x2 =3D x2(x4 – 1)
        = ;       =3D x2(x2 – 1)(x2 + 1)
        = ;       =3D x2(x – 1)(x + 1)(x2 + 1)
b. x6 – 1 =3D (x3 – 1)(x3 + 1)
        = ;      =3D (x – 1)(x2 + x + 1)(x + 1)(x2 – x + 1)
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Part= ial Fraction Decompositions
Fundamental Theorem of Algebra:
Given any real (coefficient) polynomial Q(x), = 3;
it can be written as the product of
1st degree and irreducible 2n= d deg= ree factors. 
Example:
Factor completely (over real numbers)
a. x6 – x2 =3D x2(x4 – 1)
        = ;       =3D x2(x2 – 1)(x2 + 1)
        = ;       =3D x2(x – 1)(x + 1)(x2 + 1)
b. x6 – 1 =3D (x3 – 1)(x3 + 1)
        = ;      =3D (x – 1)(x2 + x + 1)(x + 1)(x2 – x + 1)
c. x2 – 2 =3D (x – Ö2)(x + Ö2)
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Part= ial Fraction Decompositions
Reminder: To check if ax2 + b= x + c is reducible,
check b2 – 4ac. 
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Part= ial Fraction Decompositions
Reminder: To check if ax2 + b= x + c is reducible,
check b2 – 4ac. 
If b2 – 4ac > 0, it is reducibl= e.
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Part= ial Fraction Decompositions
Reminder: To check if ax2 + b= x + c is reducible,
check b2 – 4ac. 
If b2 – 4ac > 0, it is reducible.
If b2 – 4ac < 0, it is irrducib= le.
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Part= ial Fraction Decompositions
Reminder: To check if ax2 + b= x + c is reducible,
check b2 – 4ac. 
If b2 – 4ac > 0, it is reducible.
If b2 – 4ac < 0, it is irrducib= le.
Partial Fraction Decomposition Theorem:
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Part= ial Fraction Decompositions
Reminder: To check if ax2 + b= x + c is reducible,
check b2 – 4ac. 
If b2 – 4ac > 0, it is reducible.
If b2 – 4ac < 0, it is irrducib= le.
Partial Fraction Decomposition Theorem:
Given P/Q where deg P < deg Q, then
P/Q =3D F1 + F2 + .= . + Fn=  whe= re
  = ;
Fi =3D=         = ;       or 
 (ax + b)k 
(ax<= /span>2<= /span> + b= x + c)k<= /span> 
(Ai = and Bi are numbers)
Ai<= /span>x + Bi
Ai<= /span>
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Part= ial Fraction Decompositions
Reminder: To check if ax2 + b= x + c is reducible,
check b2 – 4ac. 
If b2 – 4ac > 0, it is reducible.
If b2 – 4ac < 0, it is irrducib= le.
Partial Fraction Decomposition Theorem:
Given P/Q where deg P < deg Q, then
P/Q =3D F1 + F2 + .= . + Fn=  whe= re
  = ;
Fi =3D=         = ;       or 
Ai<= /span>
 (ax + b)k 
Ai<= /span>x + Bi
(ax<= /span>2<= /span> + b= x + c)k<= /span> 
and that (ax + b)k or (ax2 + b= x + c)k=  are= factors in the factorization of Q(x).
(Ai = and Bi are numbers)
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Part= ial Fraction Decompositions
Example: a. For
P(x)
(x + 2)(x – 3)
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Part= ial Fraction Decompositions
Example: a. For
P(x)
(x + 2)(x – 3)
has two linear factors (x + 2), (x – 3),
the denominator
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Part= ial Fraction Decompositions
Example: a. For
P(x)
(x + 2)(x – 3)
has two linear factors (x + 2), (x – 3), therefoe 
P(x)
(x + 2)(x – 3)
=3D
A
(x + 2)
+
B
(x – 3 )
the denominator
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Part= ial Fraction Decompositions
Example: a. For
P(x)
(x + 2)(x – 3)
has two linear factors (x + 2), (x – 3), therefoe 
P(x)
(x + 2)(x – 3)
=3D
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
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Part= ial Fraction Decompositions
Example: a. For
P(x)
(x + 2)(x – 3)
has two linear factors (x + 2), (x – 3), therefoe 
P(x)
(x + 2)(x – 3)
=3D
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 decompositi= on are
(x + 2), (x – 3), (x – = 3)2, (x – 3)<= /span>3 and  
the denominator
, the denominator may be
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Part= ial Fraction Decompositions
Example: a. For
P(x)
(x + 2)(x – 3)
has two linear factors (x + 2), (x – 3), therefoe 
P(x)
(x + 2)(x – 3)
=3D
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 decompositi= on are
(x + 2), (x – 3), (x – = 3)2, (x – 3)<= /span>3 and  
P(x)
(x + 2)(x – 3)3 
=3D
A
(x + 2)
+
B
(x – 3)
the denominator
, the denominator may be
+
C
(x – 3)2 
+
D
(x – 3)3 
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Part= ial Fraction Decompositions
c. For
P(x)
(x + 2)(x2 + 3) 
linear factors (x + 2) and a 2nd degree irreducible <= span style=3D'position:absolute;top:26.25%;left:5.24%;width:87.26%;height:6.0%'= >factor (x2 + 3),
the denominator has one
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Part= ial Fraction Decompositions
c. For
P(x)
(x + 2)(x2 + 3) 
linear factors (x + 2) and a 2nd degree irreducible <= span style=3D'position:absolute;top:26.25%;left:5.24%;width:87.26%;height:6.0%'= >factor (x2 + 3), hence
P(x)
(x + 2)(x2 + 3) 
=3D
A
(x + 2)
+
Bx + C
(x2 + 3) 
the denominator has one
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Part= ial Fraction Decompositions
c. For
P(x)
(x + 2)(x2 + 3) 
linear factors (x + 2) and a 2nd degree irreducible <= span style=3D'position:absolute;top:26.25%;left:5.24%;width:87.26%;height:6.0%'= >factor (x2 + 3), hence
P(x)
(x + 2)(x2 + 3) 
=3D
A
(x + 2)
+
Bx + C
(x2 + 3) 
d. For
P(x)
(x + 2)2(x2 + 3)2&n= bsp;
viewed having two factors (x + 2)2, (x2 + 3)2,
the denominator has one
, the denominator may be
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Part= ial Fraction Decompositions
c. For
P(x)
(x + 2)(x2 + 3) 
linear factors (x + 2) and a 2nd degree irreducible <= span style=3D'position:absolute;top:26.25%;left:5.24%;width:87.26%;height:6.0%'= >factor (x2 + 3), hence
P(x)
(x + 2)(x2 + 3) 
=3D
A
(x + 2)
+
Bx + C
(x2 + 3) 
d. For
P(x)
(x + 2)2(x2 + 3)2&n= bsp;
viewed having two factors (x + 2)2, (x2 + 3)2, hence
the denominators in the decompositi= on are
(x + 2), (x + 2)2 , (x2 + 3), (x2 + 3)2, hence
the denominator has one
, the denominator may be
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Part= ial Fraction Decompositions
c. For
P(x)
(x + 2)(x2 + 3) 
linear factors (x + 2) and a 2nd degree irreducible <= span style=3D'position:absolute;top:26.25%;left:5.24%;width:87.26%;height:6.0%'= >factor (x2 + 3), hence
P(x)
(x + 2)(x2 + 3) 
=3D
A
(x + 2)
+
Bx + C
(x2 + 3) 
d. For
P(x)
(x + 2)2(x2 + 3)2&n= bsp;
viewed having two factors (x + 2)2, (x2 + 3)2, hence
the denominators in the decompositi= on are
(x + 2), (x + 2)2 , (x2 + 3), (x2 + 3)2, hence
=3D
A
(x + 2)
+
B
(x + 2)2 
the denominator has one
, the denominator may be
+
Cx + D
(x2 + 3) 
+
Ex + F
(x2 + 3)2&n= bsp;
P(x)
(x + 2)2(x2 + 3)2&n= bsp;
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Part= ial Fraction Decompositions
To find the exact decomposition, we use two methods:
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Part= ial Fraction Decompositions
To find the exact decomposition, we use two methods:
I. Evaluation
2. Matching coefficients
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Part= ial Fraction Decompositions
Example: Find the partial fraction&= #13;
decomposition of
To find the exact decomposition, we use two methods:
I. Evaluation
2. Matching coefficients
2x +3
(x + 2)(x – 3)
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Part= ial Fraction Decompositions
Example: Find the partial fraction&= #13;
decomposition of
We know that
(x + 2)(x – 3)
=3D
A
(x + 2)
+
B
(x – 3 )
To find the exact decomposition, we use two methods:
I. Evaluation
2. Matching coefficients
2x +3
(x + 2)(x – 3)
2x +3
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Part= ial Fraction Decompositions
Example: Find the partial fraction&= #13;
decomposition of
We know that
(x + 2)(x – 3)
=3D
A
(x + 2)
+
B
(x – 3 )
To find the exact decomposition, we use two methods:
I. Evaluation
2. Matching coefficients
2x +3
(x + 2)(x – 3)
Clear the denominator, multiply it by (x + 2)(x – 3)
2x +3
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Part= ial Fraction Decompositions
Example: Find the partial fraction&= #13;
decomposition of
We know that
(x + 2)(x – 3)
=3D
A
(x + 2)
+
B
(x – 3 )
To find the exact decomposition, we use two methods:
I. Evaluation
2. Matching coefficients
2x +3
(x + 2)(x – 3)
Clear the denominator, multiply it by (x + 2)(x – 3)
We get 2x + 3 =3D A(x – 3) + B(x + 2)
2x +3
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Part= ial Fraction Decompositions
The factors (x – 3), (x + 2) have roots at x =3D 3, x =3D -2
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Part= ial Fraction Decompositions
2x + 3 =3D A(x – 3) + B(x + 2)
The factors (x – 3), (x + 2) have roots at x =3D 3, x =3D -2
Evaluate        = ;            &n= bsp;            = ;            &n= bsp; at x =3D 3,
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Part= ial Fraction Decompositions
2x + 3 =3D A(x – 3) + B(x + 2)
The factors (x – 3), (x + 2) have roots at x =3D 3, x =3D -2
Evaluate        = ;            &n= bsp;            = ;            &n= bsp; at x =3D 3,
we get 9 =3D 0 + 5B
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Part= ial Fraction Decompositions
2x + 3 =3D A(x – 3) + B(x + 2)
The factors (x – 3), (x + 2) have roots at x =3D 3, x =3D -2
Evaluate        = ;            &n= bsp;            = ;            &n= bsp; at x =3D 3,
we get 9 =3D 0 + 5B
or  B =3D 9/5
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Part= ial Fraction Decompositions
2x + 3 =3D A(x – 3) + B(x + 2)
The factors (x – 3), (x + 2) have roots at x =3D 3, x =3D -2
Evaluate        = ;            &n= bsp;            = ;            &n= bsp; at x =3D 3,
Evaluate        = ;            &n= bsp;            = ;            &n= bsp; x =3D -2,
we get 9 =3D 0 + 5B
or  B =3D 9/5
2x + 3 =3D A(x – 3) + B(x + 2)
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Part= ial Fraction Decompositions
2x + 3 =3D A(x – 3) + B(x + 2)
The factors (x – 3), (x + 2) have roots at x =3D 3, x =3D -2
Evaluate        = ;            &n= bsp;            = ;            &n= bsp; at x =3D 3,
Evaluate        = ;            &n= bsp;            = ;            &n= bsp; x =3D -2,
we get 9 =3D 0 + 5B
or  B =3D 9/5
2x + 3 =3D A(x – 3) + B(x + 2)
we get -1 =3D -5A + 0
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