Method of Partial Fractions
edit
To decompose the rational function
P
(
x
)
Q
(
x
)
{\displaystyle {\frac {P(x)}{Q(x)}}}
:
Step 1 Use long division (if necessary) to ensure that the degree of
P
(
x
)
{\displaystyle P(x)}
is less than the degree of
Q
(
x
)
{\displaystyle Q(x)}
(see Breaking up a rational function in section 1.1 ).
Step 2 Factor Q(x) as far as possible.
Step 3 Write down the correct form for the partial fraction decomposition (see below) and solve for the constants.
To factor Q(x) we have to write it as a product of linear factors (of the form
a
x
+
b
{\displaystyle ax+b}
) and irreducible quadratic factors (of the form
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
with
b
2
−
4
a
c
<
0
{\displaystyle b^{2}-4ac<0}
).
Some of the factors could be repeated. For instance if
Q
(
x
)
=
x
3
−
6
x
2
+
9
x
{\displaystyle Q(x)=x^{3}-6x^{2}+9x}
we factor
Q
(
x
)
{\displaystyle Q(x)}
as
Q
(
x
)
=
x
(
x
2
−
6
x
+
9
)
=
x
(
x
−
3
)
(
x
−
3
)
=
x
(
x
−
3
)
2
{\displaystyle Q(x)=x(x^{2}-6x+9)=x(x-3)(x-3)=x(x-3)^{2}}
It is important that in each quadratic factor we have
b
2
−
4
a
c
<
0
{\displaystyle b^{2}-4ac<0}
, otherwise it is possible to factor that quadratic piece further. For example if
Q
(
x
)
=
x
3
−
3
x
2
+
2
x
{\displaystyle Q(x)=x^{3}-3x^{2}+2x}
then we can write
Q
(
x
)
=
x
(
x
2
−
3
x
+
2
)
=
x
(
x
−
1
)
(
x
−
2
)
{\displaystyle Q(x)=x(x^{2}-3x+2)=x(x-1)(x-2)}
We will now show how to write
P
(
x
)
Q
(
x
)
{\displaystyle {\frac {P(x)}{Q(x)}}}
as a sum of terms of the form
A
(
a
x
+
b
)
k
{\displaystyle {\frac {A}{(ax+b)^{k}}}}
and
A
x
+
B
(
a
x
2
+
b
x
+
c
)
k
{\displaystyle {\frac {Ax+B}{(ax^{2}+bx+c)^{k}}}}
Exactly how to do this depends on the factorization of
Q
(
x
)
{\displaystyle Q(x)}
and we now give four cases that can occur.
Q(x) is a product of linear factors with no repeats
edit
This means that
Q
(
x
)
=
(
a
1
x
+
b
1
)
(
a
2
x
+
b
2
)
⋯
(
a
n
x
+
b
n
)
{\displaystyle Q(x)=(a_{1}x+b_{1})(a_{2}x+b_{2})\cdots (a_{n}x+b_{n})}
where no factor is repeated and no factor is a multiple of another.
For each linear term we write down something of the form
A
(
a
x
+
b
)
{\displaystyle {\frac {A}{(ax+b)}}}
, so in total we write
P
(
x
)
Q
(
x
)
=
A
1
a
1
x
+
b
1
+
A
2
a
2
x
+
b
2
+
⋯
+
A
n
a
n
x
+
b
n
{\displaystyle {\frac {P(x)}{Q(x)}}={\frac {A_{1}}{a_{1}x+b_{1}}}+{\frac {A_{2}}{a_{2}x+b_{2}}}+\cdots +{\frac {A_{n}}{a_{n}x+b_{n}}}}
Example 1
Find
∫
1
+
x
2
(
x
+
3
)
(
x
+
5
)
(
x
+
7
)
d
x
{\displaystyle \int {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}dx}
Here we have
P
(
x
)
=
1
+
x
2
,
Q
(
x
)
=
(
x
+
3
)
(
x
+
5
)
(
x
+
7
)
{\displaystyle P(x)=1+x^{2}\ ,\ Q(x)=(x+3)(x+5)(x+7)}
and Q(x) is a product of linear factors. So we write
1
+
x
2
(
x
+
3
)
(
x
+
5
)
(
x
+
7
)
=
A
x
+
3
+
B
x
+
5
+
C
x
+
7
{\displaystyle {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}={\frac {A}{x+3}}+{\frac {B}{x+5}}+{\frac {C}{x+7}}}
Multiply both sides by the denominator
1
+
x
2
=
A
(
x
+
5
)
(
x
+
7
)
+
B
(
x
+
3
)
(
x
+
7
)
+
C
(
x
+
3
)
(
x
+
5
)
{\displaystyle 1+x^{2}=A(x+5)(x+7)+B(x+3)(x+7)+C(x+3)(x+5)}
Substitute in three values of x to get three equations for the unknown constants,
x
=
−
3
1
+
3
2
=
2
⋅
4
A
x
=
−
5
1
+
5
2
=
−
2
⋅
2
B
x
=
−
7
1
+
7
2
=
(
−
4
)
⋅
(
−
2
)
C
{\displaystyle {\begin{matrix}x=-3&1+3^{2}=2\cdot 4A\\x=-5&1+5^{2}=-2\cdot 2B\\x=-7&1+7^{2}=(-4)\cdot (-2)C\end{matrix}}}
so
A
=
5
4
,
B
=
−
13
2
,
C
=
25
4
{\displaystyle A={\tfrac {5}{4}}\ ,\ B=-{\tfrac {13}{2}}\ ,\ C={\tfrac {25}{4}}}
, and
1
+
x
2
(
x
+
3
)
(
x
+
5
)
(
x
+
7
)
=
5
4
x
+
12
−
13
2
x
+
10
+
25
4
x
+
28
{\displaystyle {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}={\frac {5}{4x+12}}-{\frac {13}{2x+10}}+{\frac {25}{4x+28}}}
We can now integrate the left hand side.
∫
1
+
x
2
(
x
+
3
)
(
x
+
5
)
(
x
+
7
)
d
x
=
5
4
ln
|
x
+
3
|
−
13
2
ln
|
x
+
5
|
+
25
4
ln
|
x
+
7
|
+
C
{\displaystyle \int {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}dx={\tfrac {5}{4}}\ln {\Big |}x+3{\Big |}-{\tfrac {13}{2}}\ln {\Big |}x+5{\Big |}+{\tfrac {25}{4}}\ln {\Big |}x+7{\Big |}+C}
Evaluate the following by the method partial fraction decomposition.
Solutions
Q(x) is a product of linear factors some of which are repeated
edit
If
(
a
x
+
b
)
{\displaystyle (ax+b)}
appears in the factorisation of
Q
(
x
)
{\displaystyle Q(x)}
k -times then instead of writing the piece
A
a
x
+
b
{\displaystyle {\frac {A}{ax+b}}}
we use the more complicated expression
A
1
a
x
+
b
+
A
2
(
a
x
+
b
)
2
+
A
3
(
a
x
+
b
)
3
+
⋯
+
A
k
(
a
x
+
b
)
k
{\displaystyle {\frac {A_{1}}{ax+b}}+{\frac {A_{2}}{(ax+b)^{2}}}+{\frac {A_{3}}{(ax+b)^{3}}}+\cdots +{\frac {A_{k}}{(ax+b)^{k}}}}
Example 2
Find
∫
d
x
(
x
+
1
)
(
x
+
2
)
2
{\displaystyle \int {\frac {dx}{(x+1)(x+2)^{2}}}}
Here
P
(
x
)
=
1
{\displaystyle P(x)=1}
and
Q
(
x
)
=
(
x
+
1
)
(
x
+
2
)
2
{\displaystyle Q(x)=(x+1)(x+2)^{2}}
We write
1
(
x
+
1
)
(
x
+
2
)
2
=
A
x
+
1
+
B
x
+
2
+
C
(
x
+
2
)
2
{\displaystyle {\frac {1}{(x+1)(x+2)^{2}}}={\frac {A}{x+1}}+{\frac {B}{x+2}}+{\frac {C}{(x+2)^{2}}}}
Multiply both sides by the denominator
1
=
A
(
x
+
2
)
2
+
B
(
x
+
1
)
(
x
+
2
)
+
C
(
x
+
1
)
{\displaystyle 1=A(x+2)^{2}+B(x+1)(x+2)+C(x+1)}
Substitute in three values of
x
{\displaystyle x}
to get 3 equations for the unknown constants,
x
=
0
1
=
2
2
A
+
2
B
+
C
x
=
−
1
1
=
A
x
=
−
2
1
=
−
C
{\displaystyle {\begin{matrix}x=0&1=2^{2}A+2B+C\\x=-1&1=A\\x=-2&1=-C\end{matrix}}}
so
A
=
1
,
B
=
−
1
,
C
=
−
1
{\displaystyle A=1\ ,\ B=-1\ ,\ C=-1}
and
1
(
x
+
1
)
(
x
+
2
)
2
=
1
x
+
1
−
1
x
+
2
−
1
(
x
+
2
)
2
{\displaystyle {\frac {1}{(x+1)(x+2)^{2}}}={\frac {1}{x+1}}-{\frac {1}{x+2}}-{\frac {1}{(x+2)^{2}}}}
We can now integrate the left hand side.
∫
d
x
(
x
+
1
)
(
x
+
2
)
2
=
ln
|
x
+
1
|
−
ln
|
x
+
2
|
+
1
x
+
2
+
C
{\displaystyle \int {\frac {dx}{(x+1)(x+2)^{2}}}=\ln \left|x+1\right|-\ln \left|x+2\right|+{\frac {1}{x+2}}+C}
We now simplify the fuction with the property of Logarithms.
ln
|
x
+
1
|
−
ln
|
x
+
2
|
+
1
x
+
2
+
C
=
ln
|
x
+
1
x
+
2
|
+
1
x
+
2
+
C
{\displaystyle \ln \left|x+1\right|-\ln \left|x+2\right|+{\frac {1}{x+2}}+C=\ln \left|{\frac {x+1}{x+2}}\right|+{\frac {1}{x+2}}+C}
3. Evaluate
∫
x
2
−
x
+
2
x
(
x
+
2
)
2
d
x
{\displaystyle \int {\frac {x^{2}-x+2}{x(x+2)^{2}}}dx}
using the method of partial fractions.
ln
|
x
(
x
+
2
)
|
2
+
4
x
+
2
+
C
{\displaystyle {\frac {\ln {\Big |}x(x+2){\Big |}}{2}}+{\frac {4}{x+2}}+C}
ln
|
x
(
x
+
2
)
|
2
+
4
x
+
2
+
C
{\displaystyle {\frac {\ln {\Big |}x(x+2){\Big |}}{2}}+{\frac {4}{x+2}}+C}
Solution
Q(x) contains some quadratic pieces which are not repeated
edit
If
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
appears we use
A
x
+
B
a
x
2
+
b
x
+
c
{\displaystyle {\frac {Ax+B}{ax^{2}+bx+c}}}
.
Evaluate the following using the method of partial fractions.
Solutions
Q(x) contains some repeated quadratic factors
edit
If
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
appears k-times then use
A
1
x
+
B
1
a
x
2
+
b
x
+
c
+
A
2
x
+
B
2
(
a
x
2
+
b
x
+
c
)
2
+
A
3
x
+
B
3
(
a
x
2
+
b
x
+
c
)
3
+
⋯
+
A
k
x
+
B
k
(
a
x
2
+
b
x
+
c
)
k
{\displaystyle {\frac {A_{1}x+B_{1}}{ax^{2}+bx+c}}+{\frac {A_{2}x+B_{2}}{(ax^{2}+bx+c)^{2}}}+{\frac {A_{3}x+B_{3}}{(ax^{2}+bx+c)^{3}}}+\cdots +{\frac {A_{k}x+B_{k}}{(ax^{2}+bx+c)^{k}}}}
Evaluate the following using the method of partial fractions.
Solution