# How do you integrate int (6-15x)/( (x-1) (x+2) (x-4)) using partial fractions?

$\textcolor{red}{\int \frac{6 - 15 x}{\left(x - 1\right) \left(x + 2\right) \left(x - 4\right)} \mathrm{dx} =}$

$\textcolor{red}{\ln \left(x - 1\right) + 2 \ln \left(x + 2\right) - 3 \ln \left(x - 4\right) + {C}_{0}}$

#### Explanation:

from the given integral, set up the equation using the variables A, B, and C

$\int \frac{6 - 15 x}{\left(x - 1\right) \left(x + 2\right) \left(x - 4\right)} \mathrm{dx} = \int \left(\frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{x - 4}\right) \mathrm{dx}$

$\frac{6 - 15 x}{\left(x - 1\right) \left(x + 2\right) \left(x - 4\right)} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{x - 4}$

$\frac{6 - 15 x}{\left(x - 1\right) \left(x + 2\right) \left(x - 4\right)} = \frac{A \left({x}^{2} - 2 x - 8\right) + B \left({x}^{2} - 5 x + 4\right) + C \left({x}^{2} + x - 2\right)}{\left(x - 1\right) \left(x + 2\right) \left(x - 4\right)}$

$\frac{6 - 15 x}{\left(x - 1\right) \left(x + 2\right) \left(x - 4\right)}$
$= \frac{\left(A + B + C\right) {x}^{2} + \left(- 2 A - 5 B + C\right) {x}^{1} + \left(- 8 A + 4 B - 2 C\right) {x}^{0}}{\left(x - 1\right) \left(x + 2\right) \left(x - 4\right)}$

The equations will be

$A + B + C = 0$
$- 2 A - 5 B + C = - 15$
$- 8 A + 4 B - 2 C = 6$

Simultaneous solution results to:

$A = 1$ and $B = 2$ and $C = - 3$

$\textcolor{red}{\int \frac{6 - 15 x}{\left(x - 1\right) \left(x + 2\right) \left(x - 4\right)} \mathrm{dx} =}$
$\textcolor{red}{\ln \left(x - 1\right) + 2 \ln \left(x + 2\right) - 3 \ln \left(x - 4\right) + {C}_{0}}$