Calculus/Integration techniques/Partial Fraction Decomposition

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	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 $3 x + 1 = A (x + 1) + B x$ . This is an equation for $x$ so 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:

Step 1 Use long division to ensure that the degree of $P (x)$ 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 $a x + b$ ) and irreducible quadratic factors (of the form $a x 2 + b x + c$ with $b 2 - 4 a c < 0$ ).

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

Q (x) = x (x 2 - 6 x + 9) = x (x - 3)(x - 3) = x (x - 3) 2 .

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

Q (x) = x (x 2 - 3 x + 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.

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

This means that $Q (x) = (a 1 x + b 1)(a 2 x + b 2)...(a n x + b n)$ where no factor is repeated and no factor is a multiple of another.

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

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

[edit] Example 1

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

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 $A = 5 / 4, B = - 13 / 2, C = 25 / 4$ , 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.

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

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

If $(a x + b)$ 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

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