# How do you integrate int x/((x-6)(x-3) dx using partial fractions?

Nov 24, 2015

$\int \frac{x}{\left(x - 6\right) \left(x - 3\right)} \mathrm{dx} = 2 \ln | x - 6 | - \ln | x - 3 | + C$

#### Explanation:

To find the partial fraction decomposition, we begin with

$\frac{x}{\left(x - 6\right) \left(x - 3\right)} = \frac{A}{x - 6} + \frac{B}{x - 3}$

Multiplying by $\left(x - 6\right) \left(x - 3\right)$ gives us

$x = A \left(x - 3\right) + B \left(x - 6\right) = \left(A + B\right) x + \left(- 3 A - 6 B\right)$

Matching respective coefficients on each side gives us the system of equations

$\left\{\begin{matrix}A + B = 1 \\ - 3 A - 6 B = 0\end{matrix}\right.$

Now all we need to do is solve for $A$ and $B$. For example, using elimination,

$\left(- 3 A - 6 B\right) + 3 \left(A + B\right) = 0 + 3 \left(1\right)$
$\implies - 3 B = 3$
$\implies B = - 1$

Then, substituting

$A - 1 = 1$
$\implies A = 2$

So the decomposition is

$\frac{x}{\left(x - 6\right) \left(x - 3\right)} = \frac{2}{x - 6} - \frac{1}{x - 3}$

Now we can do the integral relatively easily.

$\int \frac{x}{\left(x - 6\right) \left(x - 3\right)} \mathrm{dx} = \int \left(\frac{2}{x - 6} - \frac{1}{x - 3}\right) \mathrm{dx}$

$\int \left(\frac{2}{x - 6} - \frac{1}{x - 3}\right) \mathrm{dx} = 2 \int \left(\frac{1}{x - 6}\right) \mathrm{dx} - \int \left(\frac{1}{x - 3}\right) \mathrm{dx}$

A couple of simple $u$ substitutions, together with $\int \left(\frac{1}{x}\right) \mathrm{dx} = \ln | x | + C$
gives us

$2 \int \left(\frac{1}{x - 6}\right) \mathrm{dx} = 2 \ln | x - 6 | + C$
and
$- \int \left(\frac{1}{x - 3}\right) \mathrm{dx} = - \ln | x - 3 | + C$

So putting it all together, we get

$\int \frac{x}{\left(x - 6\right) \left(x - 3\right)} \mathrm{dx} = 2 \ln | x - 6 | - \ln | x - 3 | + C$