Partial Fraction Decomposition Partial Fraction Decomposition

Introduction

With

a,b,c,d \in \mathbb{R}
\frac{a}{c} + \frac{b}{d} = \frac{ad}{cd} + \frac{bc}{cd} = \frac{ad + bc}{cd}
Partial Fraction Decomposition decomposes fractions with separable denominators into separate fractions.

Multiple fractions with simpler denominators are easier to integrate than one fraction with a complex demoninator.

Example

We are solving for

A
and
B
.
\begin{align*} \int{\frac{7x+9}{(x+1)(x+2)} \, dx} &= \int{\frac{A(x+2) + B(x+1)}{(x+1)(x+2)}}\, dx \\\\ &= \int{\frac{A(x+2)}{(x+1)(x+2)} + \frac{B(x+1)}{(x+1)(x+2)} \, dx} \\\\ &= \int{\frac{A}{x+1} + \frac{B}{x+2} \, dx} \end{align*}
We can use
x=-1
and
x=-2
to find
A
and
B
.
\begin{align*} \text{When } x&=-1 \\\\ A(x+2) + B(x+1) &= 7x+9 \\\\ A(1) + B(0) &= 7(-1) + 9 \\\\ A &= 2 \\\\ \text{When } x&=-2 \\\\ A(x+2) + B(x+1) &= 7x+9 \\\\ A(0) + B(-1) &= 7(-2) + 9 \\\\ B &= 5 \\\\ 2(x+2) + 5(x+1) &= 7x + 9 \end{align*}


Plugging that into the original equation, we get
\begin{align*} \int{\frac{7x+9}{(x+1)(x+2)} \, dx} &= \int{\frac{2(x+2) + 5(x+1)}{(x+1)(x+2)} \, dx} \\\\ &= \int{\frac{2}{x+1} + \frac{5}{x+2} \, dx} \\\\ &= 2\ln|x+1| + 5\ln|x+2| + C \end{align*}

Example #2

Partial fraction decomposition is linear when the results have linear denominators.

For this course, if you find yourself decomposing into non-linear denominators, such as

(x+1)^2
or
\sqrt[3]{x}
, you are probably doing it wrong.

Our second integral is
\int{\frac{4x^2 + 7x + 1}{(x+3)(x-2)(x+2)} \, dx} = \int{\frac{A}{x+3} + \frac{B}{x-2} + \frac{C}{x+2} \, dx} \\ \text{} \\ A(x-2)(x+2) + B(x+3)(x+2) + C(x+3)(x-2) = 4x^2 + 7x + 1 \\ \text{} \\ \text{When } x=-3 \\ \text{} \\ A(-5)(-1) + B(0) + C(0) = 36 - 21 + 1 \\ \text{} \\ A = \frac{16}{5} \\ \text{} \\ \text{When } x=2 \\ \text{} \\ A(0) + B(5)(4) + C(0) = 16 + 14 + 1 \\ \text{} \\ B = \frac{31}{20} \\ \text{} \\ \text{When } x=-2 \\ \text{} \\ A(0) + B(0) + C(1)(-4) = 16 - 14 + 1 \\ \text{} \\ C = \frac{-3}{4} \\ \text{} \\
\begin{align*} \int{\frac{4x^2 + 7x + 1}{(x+3)(x-2)(x+2)} \, dx} &= \int{\frac{\frac{16}{5}}{x+3} + \frac{\frac{31}{20}}{x-2} + \frac{\frac{-3}{4}}{x+2} \, dx} \\ \text{} \\ &= \frac{16}{5}\ln|x+3| + \frac{31}{20}\ln|x-2| - \frac{3}{4}\ln|x+2| + C \end{align*}