Calculus/Integration techniques/Partial Fraction Decomposition

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

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

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 $Q(x)$ (see Breaking up a rational function in section Template:Calculus/map page).
• 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 $a{x}^2+bx+c$ 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 $\frac{Ax+B}{(a{x}^2+bx+c{)}^k}$

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

$\frac{P(x)}{Q(x)}=\frac{A_1}{a_{1}x+b_1}+\frac{A_2}{a_{2}x+b_2}+\cdots +\frac{A_n}{a_{n}x+b_n}$

Template:ExampleRobox 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,

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

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

$\int \frac{1+x^2}{(x+3)(x+5)(x+7)}dx=\frac{5}{4}\ln |x+3|-\frac{13}{2}\ln |x+5|+\frac{25}{4}\ln |x+7|+C$ Template:Robox/Close

Exercises

Evaluate the following by the method partial fraction decomposition. Template:Question-answer

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

$\frac{A_1}{ax+b}+\frac{A_2}{(ax+b{)}^2}+\frac{A_3}{(ax+b{)}^3}+\cdots +\frac{A_k}{(ax+b{)}^k}$

Template:ExampleRobox 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,

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

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.

$\int \frac{dx}{(x+1)(x+2{)}^2}=\ln |x+1|-\ln |x+2|+\frac{1}{x+2}+C$

We now simplify the fuction with the property of Logarithms.

$\ln |x+1|-\ln |x+2|+\frac{1}{x+2}+C=\ln \left|\frac{x+1}{x+2}\right|+\frac{1}{x+2}+C$

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

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

Exercises

Evaluate the following using the method of partial fractions. Template:Question-answer

Q(x) contains some repeated quadratic factors

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

$\frac{A_{1}x+B_1}{a{x}^2+bx+c}+\frac{A_{2}x+B_2}{(a{x}^2+bx+c{)}^2}+\frac{A_{3}x+B_3}{(a{x}^2+bx+c{)}^3}+\cdots +\frac{A_{k}x+B_k}{(a{x}^2+bx+c{)}^k}$

Exercise

Evaluate the following using the method of partial fractions. Template:Question-answer Template:Noprint