# How do you find int (x - 4 ) /((x-1)(x+2)(x-3)) dx using partial fractions?

Apr 10, 2016

$\int \frac{x - 4}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)} \mathrm{dx}$

$= \int \frac{1}{2 \left(x - 1\right)} - \frac{2}{5 \left(x + 2\right)} - \frac{1}{10 \left(x - 3\right)} \mathrm{dx}$

$= \frac{1}{2} \ln \left\mid x - 1 \right\mid - \frac{2}{5} \ln \left\mid x + 2 \right\mid - \frac{1}{10} \ln \left\mid x - 3 \right\mid + C$

#### Explanation:

Since the denominator is already factored into distinct linear factors, we are looking for a partial fraction decompositionof the form:

$\frac{x - 4}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{x - 3}$

$= \frac{A \left(x + 2\right) \left(x - 3\right) + B \left(x - 1\right) \left(x - 3\right) + C \left(x - 1\right) \left(x + 2\right)}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)}$

$= \frac{A \left({x}^{2} - x - 6\right) + B \left({x}^{2} - 4 x + 3\right) + C \left({x}^{2} + x - 2\right)}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)}$

$= \frac{\left(A + B + C\right) {x}^{2} + \left(- A - 4 B + C\right) x + \left(- 6 A + 3 B - 2 C\right)}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)}$

Equating coefficients, we get the system of simulataneous linear equations:

$\left\{\begin{matrix}A + B + C = 0 \\ - A - 4 B + C = 1 \\ - 6 A + 3 B - 2 C = - 4\end{matrix}\right.$

Add multiples of the first equation to the second and third equations to get:

$\left\{\begin{matrix}A + B + C = 0 \\ - 3 B + 2 C = 1 \\ 9 B + 4 C = - 4\end{matrix}\right.$

Add three times the second equation to the third equation to get:

$\left\{\begin{matrix}A + B + C = 0 \\ - 3 B + 2 C = 1 \\ 10 C = - 1\end{matrix}\right.$

Divide the third equation by $10$ to get:

$\left\{\begin{matrix}A + B + C = 0 \\ - 3 B + 2 C = 1 \\ C = - \frac{1}{10}\end{matrix}\right.$

Subtract multiples of the third equation from the other equations to get:

$\left\{\begin{matrix}A + B = \frac{1}{10} \\ - 3 B = \frac{6}{5} \\ C = - \frac{1}{10}\end{matrix}\right.$

Divide the second equation by $- 3$ to get:

$\left\{\begin{matrix}A + B = \frac{1}{10} \\ B = - \frac{2}{5} \\ C = - \frac{1}{10}\end{matrix}\right.$

Subtract the second equation from the first to get:

$\left\{\begin{matrix}A = \frac{1}{2} \\ B = - \frac{2}{5} \\ C = - \frac{1}{10}\end{matrix}\right.$

So:

$\frac{x - 4}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)} = \frac{1}{2 \left(x - 1\right)} - \frac{2}{5 \left(x + 2\right)} - \frac{1}{10 \left(x - 3\right)}$

So:

$\int \frac{x - 4}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)} \mathrm{dx}$

$= \int \frac{1}{2 \left(x - 1\right)} - \frac{2}{5 \left(x + 2\right)} - \frac{1}{10 \left(x - 3\right)} \mathrm{dx}$

$= \frac{1}{2} \ln \left\mid x - 1 \right\mid - \frac{2}{5} \ln \left\mid x + 2 \right\mid - \frac{1}{10} \ln \left\mid x - 3 \right\mid + C$