# How do you use partial fraction decomposition to decompose the fraction to integrate (13x) / (6x^2 + 5x - 6)?

Aug 15, 2015

Using a fraction whose denominator is quadratic, you need to check whether or not it is factorable.

$6 {x}^{2} + 5 x - 6 = \left(2 x + 3\right) \left(3 x - 2\right)$

Since it is, we only need to write out the following:

$\int \frac{13 x}{\left(2 x + 3\right) \left(3 x - 2\right)} \mathrm{dx} = \int \frac{A}{2 x + 3} + \frac{B}{3 x - 2} \mathrm{dx}$

To find what $A$ and $B$ are, you could start by cross-multiplying. For now, ignore the integral sign and $\mathrm{dx}$.

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

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

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

Thus, we have the equations:

$- 2 A + 3 B = 0 \implies B = \frac{2}{3} A$

$3 A + 2 B = 13 \implies 3 A + \frac{4}{3} A = 13$
$\frac{13}{3} A = 13$
$\therefore \textcolor{g r e e n}{A = 3} \implies \textcolor{g r e e n}{B = 2}$

At this point we are pretty much done.

$= \int \frac{3}{2 x + 3} + \frac{2}{3 x - 2} \mathrm{dx}$

$= \textcolor{b l u e}{\frac{3}{2} \ln | 2 x + 3 | + \frac{2}{3} \ln | 3 x - 2 | + C}$