# How do you integrate int frac{2x-4}{(x-4)(x+3)(x-6)} dx using partial fractions?

Dec 28, 2015

$\int \frac{2 x - 4}{\left(x - 4\right) \left(x + 3\right) \left(x - 6\right)} \mathrm{dx}$

$= - \frac{2}{7} \ln | x - 4 | - \frac{10}{63} \ln | x + 3 | + \frac{4}{9} \ln | x - 6 | + C$

#### Explanation:

Applying partial fraction decomposition:

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

$\implies 2 x - 4 = A \left(x + 3\right) \left(x - 6\right) + B \left(x - 4\right) \left(x - 6\right) + C \left(x - 4\right) \left(x + 3\right)$

$\implies 2 x - 4 = \left(A + B + C\right) {x}^{2} + \left(- 3 A - 10 B - C\right) x + \left(- 18 A + 24 B - 12 C\right)$

Equating equivalent coefficients gives us the system

$\left\{\begin{matrix}A + B + C = 0 \\ - 3 A - 10 B - C = 2 \\ - 18 A + 24 B - 12 C = 4\end{matrix}\right.$

Solving, we get

$\left\{\begin{matrix}A = - \frac{2}{7} \\ B = - \frac{10}{63} \\ C = \frac{4}{9}\end{matrix}\right.$

Thus, substituting back in,

$\frac{2 x - 4}{\left(x - 4\right) \left(x + 3\right) \left(x - 6\right)} = - \frac{\frac{2}{7}}{x - 4} - \frac{\frac{10}{63}}{x + 3} + \frac{\frac{4}{9}}{x - 6}$

and so, integrating,

$\int \frac{2 x - 4}{\left(x - 4\right) \left(x + 3\right) \left(x - 6\right)} \mathrm{dx}$

$= \int \left(- \frac{\frac{2}{7}}{x - 4} - \frac{\frac{10}{63}}{x + 3} + \frac{\frac{4}{9}}{x - 6}\right) \mathrm{dx}$

$= - \left(\frac{2}{7}\right) \int \frac{1}{x - 4} \mathrm{dx} - \left(\frac{10}{63}\right) \int \frac{1}{x + 3} \mathrm{dx}$
$+ \left(\frac{4}{9}\right) \int \frac{1}{x - 6} \mathrm{dx}$

$= - \frac{2}{7} \ln | x - 4 | - \frac{10}{63} \ln | x + 3 | + \frac{4}{9} \ln | x - 6 | + C$

Dec 28, 2015

$\int \frac{2 x - 4}{\left(x - 4\right) \left(x + 3\right) \left(x - 6\right)} \mathrm{dx}$
$= - \frac{2}{7} \ln | x - 4 | - \frac{10}{63} \ln | x + 3 | + \frac{4}{9} \ln | x - 6 | + C$

where $C$ is the constant of integration

#### Explanation:

First, you need to write out the partial fractions. The denominator has already been factorized for you.

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

Where $A$, $B$ and $C$ are constants to be determined. Note that the sign $\equiv$ means that the equality holds true for all possible values of $x$. Get rid of the denominators by multiplying both sides with $\left(x - 4\right) \left(x + 3\right) \left(x - 6\right)$.

$2 x - 4 \equiv A \left(x + 3\right) \left(x - 6\right) + B \left(x - 4\right) \left(x - 6\right) + C \left(x - 4\right) \left(x + 3\right)$

When $x = 4$

$2 \left(4\right) - 4 = A \left(4 + 3\right) \left(4 - 6\right)$

$A = - \frac{2}{7}$

When $x = - 3$

$2 \left(- 3\right) - 4 = B \left(- 3 - 4\right) \left(- 3 - 6\right)$

$B = - \frac{10}{63}$

When $x = 6$

$2 \left(6\right) - 4 = C \left(6 - 4\right) \left(6 + 3\right)$

$C = \frac{4}{9}$

Therefore,

$\frac{2 x - 4}{\left(x - 4\right) \left(x + 3\right) \left(x - 6\right)} \equiv - \frac{\frac{2}{7}}{x - 4} - \frac{\frac{10}{63}}{x + 3} + \frac{\frac{4}{9}}{x - 6}$.

Now, we proceed with the integration.

$\int \frac{2 x - 4}{\left(x - 4\right) \left(x + 3\right) \left(x - 6\right)} \mathrm{dx} = \int \left(- \frac{\frac{2}{7}}{x - 4} - \frac{\frac{10}{63}}{x + 3} + \frac{\frac{4}{9}}{x - 6}\right) \mathrm{dx}$

$= - \frac{2}{7} \int \frac{1}{x - 4} \mathrm{dx} - \frac{10}{63} \int \frac{1}{x + 3} \mathrm{dx} + \frac{4}{9} \int \frac{1}{x - 6} \mathrm{dx}$

$= - \frac{2}{7} \ln | x - 4 | - \frac{10}{63} \ln | x + 3 | + \frac{4}{9} \ln | x - 6 | + C$,

where $C$ is the constant of integration.