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


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

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 $$Q(x)$$ (see Breaking up a rational function in section ).
 * 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 $$Q(x)=(a_1x+b_1)(a_2x+b_2)\cdots(a_nx+b_n)$$ 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


 * $$\frac{P(x)}{Q(x)}=\frac{A_1}{a_1x+b_1}+\frac{A_2}{a_2x+b_2}+\cdots+\frac{A_n}{a_nx+b_n}$$

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,

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

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.

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

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

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,

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

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

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.

Q(x) contains some repeated quadratic factors
If $$ax^2+bx+c$$ appears k-times then use
 * $$\frac{A_1x+B_1}{ax^2+bx+c}+\frac{A_2x+B_2}{(ax^2+bx+c)^2}+\frac{A_3x+B_3}{(ax^2+bx+c)^3}+\cdots+\frac{A_kx+B_k}{(ax^2+bx+c)^k}$$

Exercise
Evaluate the following using the method of partial fractions.