Partial Fractions

Suppose we want to find $\int {\frac {3x+1}{x^{2}+x}}dx$ . One way to do this is to simplify the integrand by finding constants $A$ and $B$ so that

{\frac {3x+1}{x^{2}+x}}={\frac {3x+1}{x(x+1)}}={\frac {A}{x}}+{\frac {B}{x+1}}

.

This can be done by cross multiplying the fraction which gives

{\frac {3x+1}{x(x+1)}}={\frac {A(x+1)+Bx}{x(x+1)}}

As both sides have the same denominator we must have

3x+1=A(x+1)+Bx

This is an equation for $x$ so it must hold whatever value $x$ is. If we put in $x=0$ we get $A=1$ and putting $x=-1$ gives $=-B=-2$ so $B=2$ . So we see that

{\frac {3x+1}{x^{2}+x}}={\frac {1}{x}}+{\frac {2}{x+1}}

Returning to the original integral

$\int {\frac {3x+1}{x^{2}+x}}dx$	$=\int {\frac {dx}{x}}+\int {\frac {2}{x+1}}dx$
	$=\int {\frac {dx}{x}}+2\int {\frac {dx}{x+1}}$
	$=\ln \|x\|+2\ln {\Big \|}x+1{\Big \|}+C$

Rewriting the integrand as a sum of simpler fractions has allowed us to reduce the initial integral to a sum of simpler integrals. In fact this method works to integrate any rational function.

Method of Partial Fractions

To decompose the rational function ${\frac {P(x)}{Q(x)}}$ :

Step 1 Use long division (if necessary) to ensure that the degree of $P(x)$ is less than the degree of $Q(x)$ .

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 $ax+b$ ) and irreducible quadratic factors (of the form $ax^{2}+bx+c$ with $b^{2}-4ac<0$ ).

Some of the factors could be repeated. For instance if $Q(x)=x^{3}-6x^{2}+9x$ we factor $Q(x)$ as

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 $b^{2}-4ac<0$ , otherwise it is possible to factor that quadratic piece further. For example if $Q(x)=x^{3}-3x^{2}+2x$ then we can write

Q(x)=x(x^{2}-3x+2)=x(x-1)(x-2)

We will now show how to write ${\frac {P(x)}{Q(x)}}$ as a sum of terms of the form

{\frac {A}{(ax+b)^{k}}}

and

{\frac {Ax+B}{(ax^{2}+bx+c)^{k}}}

Exactly how to do this depends on the factorization of $Q(x)$ and we now give four cases that can occur.

Q(x) is a product of linear factors with no repeats

This means that $Q(x)=(a_{1}x+b_{1})(a_{2}x+b_{2})\cdots (a_{n}x+b_{n})$ where no factor is repeated and no factor is a multiple of another.

For each linear term we write down something of the form ${\frac {A}{(ax+b)}}$ , so in total we write

{\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 Problem

Find $\int {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}dx$

Here we have $P(x)=1+x^{2}\ ,\ Q(x)=(x+3)(x+5)(x+7)$ and Q(x) is a product of linear factors. So we write

${\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}={\frac {A}{x+3}}+{\frac {B}{x+5}}+{\frac {C}{x+7}}$

Multiply both sides by the denominator

$1+x^{2}=A(x+5)(x+7)+B(x+3)(x+7)+C(x+3)(x+5)$

Substitute in three values of x to get three equations for the unknown constants,

${\begin{matrix}x=-3&1+3^{2}=2\cdot 4A\\x=-5&1+5^{2}=-2\cdot 2B\\x=-7&1+7^{2}=(-4)\cdot (-2)C\end{matrix}}$

so $A={\tfrac {5}{4}}\ ,\ B=-{\tfrac {13}{2}}\ ,\ C={\tfrac {25}{4}}$ , and

${\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}={\frac {5}{4x+12}}-{\frac {13}{2x+10}}+{\frac {25}{4x+28}}$

We can now integrate the left hand side.

$\int {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}dx={\tfrac {5}{4}}\ln {\Big |}x+3{\Big |}-{\tfrac {13}{2}}\ln {\Big |}x+5{\Big |}+{\tfrac {25}{4}}\ln {\Big |}x+7{\Big |}+C$

Exercises

Evaluate the following by the method partial fraction decomposition. 1. $\int {\frac {2x+11}{(x+6)(x+5)}}dx$

2. $\int {\frac {7x^{2}-5x+6}{(x-1)(x-3)(x-7)}}dx$

Q(x) is a product of linear factors some of which are repeated

If $(ax+b)$ appears in the factorisation of $Q(x)$ k-times then instead of writing the piece ${\frac {A}{ax+b}}$ we use the more complicated expression

${\frac {A_{1}}{ax+b}}+{\frac {A_{2}}{(ax+b)^{2}}}+{\frac {A_{3}}{(ax+b)^{3}}}+\cdots +{\frac {A_{k}}{(ax+b)^{k}}}$

Example 2

Find $\int {\frac {dx}{(x+1)(x+2)^{2}}}$

Here $P(x)=1$ and $Q(x)=(x+1)(x+2)^{2}$ We write

${\frac {1}{(x+1)(x+2)^{2}}}={\frac {A}{x+1}}+{\frac {B}{x+2}}+{\frac {C}{(x+2)^{2}}}$

Multiply both sides by the denominator $1=A(x+2)^{2}+B(x+1)(x+2)+C(x+1)$

Substitute in three values of $x$ to get 3 equations for the unknown constants,

${\begin{matrix}x=0&1=2^{2}A+2B+C\\x=-1&1=A\\x=-2&1=-C\end{matrix}}$

so $A=1\ ,\ B=-1\ ,\ C=-1$ and

${\frac {1}{(x+1)(x+2)^{2}}}={\frac {1}{x+1}}-{\frac {1}{x+2}}-{\frac {1}{(x+2)^{2}}}$

We can now integrate the left hand side.

\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 \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

Exercise

3. Evaluate $\int {\frac {x^{2}-x+2}{x(x+2)^{2}}}dx$ using the method of partial fractions.

Q(x) contains some quadratic pieces which are not repeated

If $ax^{2}+bx+c$ appears we use ${\frac {Ax+B}{ax^{2}+bx+c}}$ .

Exercises

Evaluate the following using the method of partial fractions. 4. $\int {\frac {2}{(x+2)(x^{2}+3)}}dx$

5. $\int {\frac {dx}{(x+2)(x^{2}+2)}}$

Q(x) contains some repeated quadratic factors

If $ax^{2}+bx+c$ appears k-times then use

{\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}}}

Exercise

Evaluate the following using the method of partial fractions.

6. $\int {\frac {dx}{(x-1)(x^{2}+1)^{2}}}$

Exercies Solutions

$\ln {\Big |}x+6{\Big |}+\ln {\Big |}x+5{\Big |}+C$
${\tfrac {2}{3}}\ln {\Big |}x-1{\Big |}-{\tfrac {27}{4}}\ln {\Big |}x-3{\Big |}+{\tfrac {157}{12}}\ln {\Big |}x-7{\Big |}+C$
${\frac {\ln {\Big |}x(x+2){\Big |}}{2}}+{\frac {4}{x+2}}+C$
$\ln \left({\sqrt[{7}]{\frac {(x+2)^{2}}{x^{2}+3}}}\right)+{\frac {4\arctan {\Big (}{\tfrac {x}{\sqrt {3}}}{\Big )}}{7{\sqrt {3}}}}+C$
$\ln \left({\sqrt[{12}]{\frac {(x+2)^{2}}{x^{2}+2}}}\right)+{\frac {{\sqrt {2}}\arctan {\Big (}{\tfrac {x}{\sqrt {2}}}{\Big )}}{6}}+C$
${\frac {1-x}{4(x^{2}+1)}}+{\tfrac {1}{8}}\ln \left({\frac {(x-1)^{2}}{x^{2}+1}}\right)-{\frac {\arctan(x)}{2}}+C$