Calculus/Integration techniques/Partial Fraction Decomposition

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

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

This can be done by cross multiplying the fraction which gives

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

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 1 = A and putting x = − 1 gives − 2 = − B 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

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

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 to ensure that the degree of P(x) is less than the degree of 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 ax + b) and irreducible quadratic factors (of the form ax² + bx + c with b² − 4ac < 0).

Some of the factors could be repeated. For instance if Q(x) = x³ − 6x² + 9x we factor Q(x) as

Q(x) = x(x² − 6x + 9) = x(x − 3)(x − 3) = x(x − 3)².

It is important that in each quadratic factor we have b² − 4ac < 0, otherwise it is possible to factor that quadratic piece further. For example if Q(x) = x³ − 3x² − 2x then we can write

Q(x) = x(x² − 3x + 2) = x(x − 1)(x + 2)

We will now show how to write P(x) / Q(x) as a sum of terms of the form

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

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) = (a₁x + b₁)(a₂x + b₂)...(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 ${A \over (ax+b)}$, so in total we write

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

Example 1 Find $\int {1+x^2 \over (x+3)(x+5)(x+7)}dx$ Here we have P(x) = 1 + x2,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 + x2 = 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 = 5 / 4,B = − 13 / 2,C = 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. $\int \frac{1+x^2 \, dx}{(x+3)(x+5)(x+7)}= \frac{5}{4} \ln|x+3| - \frac{13}{2}\ln|x+5|+ \frac{25}{4}\ln|x+7|+ C$

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 ${A \over (ax+b)}$ we use the more complicated expression

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

Example 2 Find $\int {1 \over (x+1)(x+2)^2}dx$ 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. $\int \frac{1}{(x+1)(x+2)^2} dx= \ln \frac{x+1}{x+2} + \frac{1}{x+2} +C$

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 (ax² + bx + c) appears we use ${Ax+B \over (ax^2+bx+c)}.$

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 (ax² + bx + c) appears k-times then use

${A_1x+B_1 \over (ax^2+bx+c)} + {A_2x+B_2 \over (ax^2+bx+c)^2} + {A_3x+B_3 \over (ax^2+bx+c)^3} + \cdots + {A_kx+B_k \over (ax^2+bx+c)^k}$

Exercise

Evaluate the following using the method of partial fractions.

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