How do you integrate int (x-9)/((x+3)(x-6)(x+4))  using partial fractions?

Dec 21, 2015

You need to decompose $\frac{x - 9}{\left(x + 3\right) \left(x - 6\right) \left(x + 4\right)}$ as a partial fraction.

You're looking for $a , b , c \in \mathbb{R}$ such that $\frac{x - 9}{\left(x + 3\right) \left(x - 6\right) \left(x + 4\right)} = \frac{a}{x + 3} + \frac{b}{x - 6} + \frac{c}{x + 4}$. I'm gonna show you how to find $a$ only, because $b$ and $c$ are to be found in the exact same way.

You multiply both sides by $x + 3$, this will make it disappear from the denominator of the left side and make it appear next to $b$ and $c$.

$\frac{x - 9}{\left(x + 3\right) \left(x - 6\right) \left(x + 4\right)} = \frac{a}{x + 3} + \frac{b}{x - 6} + \frac{c}{x + 4} \iff \frac{x - 9}{\left(x - 6\right) \left(x + 4\right)} = a + \frac{b \left(x + 3\right)}{x - 6} + \frac{c \left(x + 3\right)}{x + 4}$. You evaluate this at $x - 3$ in order to make $b$ and $c$ disappear and find $a$.

$x = - 3 \iff \frac{12}{9} = \frac{4}{3} = a$. You do the same for $b$ and $c$, except that you multiply both sides by their respective denominators, and you will find out that $b = - \frac{1}{30}$ and $c = - \frac{13}{10}$.

It means we now have to integrate $\frac{4}{3} \int \frac{\mathrm{dx}}{x + 3} - \frac{1}{30} \int \frac{\mathrm{dx}}{x - 6} - \frac{13}{10} \int \frac{\mathrm{dx}}{x + 4} = \frac{4}{3} \ln \left\mid x + 3 \right\mid - \frac{1}{30} \ln \left\mid x - 6 \right\mid - \frac{13}{10} \ln \left\mid x + 4 \right\mid$