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*}