# Calculus/Integration techniques/Partial Fraction Decomposition

 ← Integration techniques/Trigonometric Integrals Calculus Integration techniques/Tangent Half Angle → Integration techniques/Partial Fraction Decomposition

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

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

This can be done by cross multiplying the fraction which gives

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

As both sides have the same denominator we must have

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

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

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

Returning to the original integral

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

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 ${\displaystyle {\frac {P(x)}{Q(x)}}}$:

• Step 1 Use long division (if necessary) to ensure that the degree of ${\displaystyle P(x)}$ is less than the degree of ${\displaystyle Q(x)}$ (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 ${\displaystyle ax+b}$) and irreducible quadratic factors (of the form ${\displaystyle ax^{2}+bx+c}$ with ${\displaystyle b^{2}-4ac<0}$).

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

${\displaystyle Q(x)=x(x^{2}-6x+9)=x(x-3)(x-3)=x(x-3)^{2}}$

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

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

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

${\displaystyle {\frac {A}{(ax+b)^{k}}}}$ and ${\displaystyle {\frac {Ax+B}{(ax^{2}+bx+c)^{k}}}}$

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

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

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

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

${\displaystyle {\frac {P(x)}{Q(x)}}={\frac {A_{1}}{a_{1}x+b_{1}}}+{\frac {A_{2}}{a_{2}x+b_{2}}}+\cdots +{\frac {A_{n}}{a_{n}x+b_{n}}}}$
 Example 1 Find ${\displaystyle \int {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}dx}$ Here we have ${\displaystyle P(x)=1+x^{2}\ ,\ Q(x)=(x+3)(x+5)(x+7)}$ and Q(x) is a product of linear factors. So we write ${\displaystyle {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}={\frac {A}{x+3}}+{\frac {B}{x+5}}+{\frac {C}{x+7}}}$ Multiply both sides by the denominator ${\displaystyle 1+x^{2}=A(x+5)(x+7)+B(x+3)(x+7)+C(x+3)(x+5)}$ Substitute in three values of x to get three equations for the unknown constants, ${\displaystyle {\begin{matrix}x=-3&1+3^{2}=2\cdot 4A\\x=-5&1+5^{2}=-2\cdot 2B\\x=-7&1+7^{2}=(-4)\cdot (-2)C\end{matrix}}}$ so ${\displaystyle A={\tfrac {5}{4}}\ ,\ B=-{\tfrac {13}{2}}\ ,\ C={\tfrac {25}{4}}}$ , and ${\displaystyle {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}={\frac {5}{4x+12}}-{\frac {13}{2x+10}}+{\frac {25}{4x+28}}}$ We can now integrate the left hand side. ${\displaystyle \int {\frac {1+x^{2}}{(x+3)(x+5)(x+7)}}dx={\tfrac {5}{4}}\ln {\Big |}x+3{\Big |}-{\tfrac {13}{2}}\ln {\Big |}x+5{\Big |}+{\tfrac {25}{4}}\ln {\Big |}x+7{\Big |}+C}$

#### Exercises

Evaluate the following by the method partial fraction decomposition.

1. ${\displaystyle \int {\frac {2x+11}{(x+6)(x+5)}}dx}$
${\displaystyle \ln {\Big |}x+6{\Big |}+\ln {\Big |}x+5{\Big |}+C}$
${\displaystyle \ln {\Big |}x+6{\Big |}+\ln {\Big |}x+5{\Big |}+C}$
2. ${\displaystyle \int {\frac {7x^{2}-5x+6}{(x-1)(x-3)(x-7)}}dx}$
${\displaystyle {\tfrac {2}{3}}\ln {\Big |}x-1{\Big |}-{\tfrac {27}{4}}\ln {\Big |}x-3{\Big |}+{\tfrac {157}{12}}\ln {\Big |}x-7{\Big |}+C}$
${\displaystyle {\tfrac {2}{3}}\ln {\Big |}x-1{\Big |}-{\tfrac {27}{4}}\ln {\Big |}x-3{\Big |}+{\tfrac {157}{12}}\ln {\Big |}x-7{\Big |}+C}$

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

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

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

 Example 2 Find ${\displaystyle \int {\frac {dx}{(x+1)(x+2)^{2}}}}$ Here ${\displaystyle P(x)=1}$ and ${\displaystyle Q(x)=(x+1)(x+2)^{2}}$ We write ${\displaystyle {\frac {1}{(x+1)(x+2)^{2}}}={\frac {A}{x+1}}+{\frac {B}{x+2}}+{\frac {C}{(x+2)^{2}}}}$ Multiply both sides by the denominator ${\displaystyle 1=A(x+2)^{2}+B(x+1)(x+2)+C(x+1)}$ Substitute in three values of ${\displaystyle x}$ to get 3 equations for the unknown constants, ${\displaystyle {\begin{matrix}x=0&1=2^{2}A+2B+C\\x=-1&1=A\\x=-2&1=-C\end{matrix}}}$ so ${\displaystyle A=1\ ,\ B=-1\ ,\ C=-1}$ and ${\displaystyle {\frac {1}{(x+1)(x+2)^{2}}}={\frac {1}{x+1}}-{\frac {1}{x+2}}-{\frac {1}{(x+2)^{2}}}}$ We can now integrate the left hand side. ${\displaystyle \int {\frac {dx}{(x+1)(x+2)^{2}}}=\ln \left|x+1\right|-\ln \left|x+2\right|+{\frac {1}{x+2}}+C}$ We now simplify the fuction with the property of Logarithms. ${\displaystyle \ln \left|x+1\right|-\ln \left|x+2\right|+{\frac {1}{x+2}}+C=\ln \left|{\frac {x+1}{x+2}}\right|+{\frac {1}{x+2}}+C}$

#### Exercise

3. Evaluate ${\displaystyle \int {\frac {x^{2}-x+2}{x(x+2)^{2}}}dx}$ using the method of partial fractions.
${\displaystyle {\frac {\ln {\Big |}x(x+2){\Big |}}{2}}+{\frac {4}{x+2}}+C}$
${\displaystyle {\frac {\ln {\Big |}x(x+2){\Big |}}{2}}+{\frac {4}{x+2}}+C}$

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

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

#### Exercises

Evaluate the following using the method of partial fractions.

4. ${\displaystyle \int {\frac {2}{(x+2)(x^{2}+3)}}dx}$
${\displaystyle \ln \left({\sqrt[{7}]{\frac {(x+2)^{2}}{x^{2}+3}}}\right)+{\frac {4\arctan {\Big (}{\tfrac {x}{\sqrt {3}}}{\Big )}}{7{\sqrt {3}}}}+C}$
${\displaystyle \ln \left({\sqrt[{7}]{\frac {(x+2)^{2}}{x^{2}+3}}}\right)+{\frac {4\arctan {\Big (}{\tfrac {x}{\sqrt {3}}}{\Big )}}{7{\sqrt {3}}}}+C}$
5. ${\displaystyle \int {\frac {dx}{(x+2)(x^{2}+2)}}}$
${\displaystyle \ln \left({\sqrt[{12}]{\frac {(x+2)^{2}}{x^{2}+2}}}\right)+{\frac {{\sqrt {2}}\arctan {\Big (}{\tfrac {x}{\sqrt {2}}}{\Big )}}{6}}+C}$
${\displaystyle \ln \left({\sqrt[{12}]{\frac {(x+2)^{2}}{x^{2}+2}}}\right)+{\frac {{\sqrt {2}}\arctan {\Big (}{\tfrac {x}{\sqrt {2}}}{\Big )}}{6}}+C}$

### Q(x) contains some repeated quadratic factors

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

${\displaystyle {\frac {A_{1}x+B_{1}}{ax^{2}+bx+c}}+{\frac {A_{2}x+B_{2}}{(ax^{2}+bx+c)^{2}}}+{\frac {A_{3}x+B_{3}}{(ax^{2}+bx+c)^{3}}}+\cdots +{\frac {A_{k}x+B_{k}}{(ax^{2}+bx+c)^{k}}}}$

#### Exercise

Evaluate the following using the method of partial fractions.

6. ${\displaystyle \int {\frac {dx}{(x-1)(x^{2}+1)^{2}}}}$
${\displaystyle {\frac {1-x}{4(x^{2}+1)}}+{\tfrac {1}{8}}\ln \left({\frac {(x-1)^{2}}{x^{2}+1}}\right)-{\frac {\arctan(x)}{2}}+C}$
${\displaystyle {\frac {1-x}{4(x^{2}+1)}}+{\tfrac {1}{8}}\ln \left({\frac {(x-1)^{2}}{x^{2}+1}}\right)-{\frac {\arctan(x)}{2}}+C}$
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