# How do you integrate int (3x-4)/((x-1)(x+3)(x-6))  using partial fractions?

$\frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{x - 6}$
Cover up rule or simultaneously solving by equating coefficients yields $A = \frac{1}{20}$, $B = - \frac{13}{36}$and $C = \frac{14}{45}$
$\int \frac{3 x - 4}{\left(x - 1\right) \left(x + 3\right) \left(x - 6\right)} \mathrm{dx}$
$= \frac{1}{20} \int \frac{1}{x - 1} \mathrm{dx} - \frac{13}{36} \int \frac{1}{x + 3} \mathrm{dx} + \frac{14}{45} \int \left(\frac{1}{x} - 6\right) \mathrm{dx}$
$= \frac{1}{20} \ln | x - 1 | - \frac{13}{36} \ln | x + 3 | + \frac{14}{45} \ln | x - 6 | + c$