# How do you integrate int (13x) / (6x^2 + 5x - 6) using partial fractions?

Jun 14, 2016

$\frac{2}{3} \ln \left(\left\mid 3 x - 2 \right\mid\right) + \frac{3}{2} \ln \left(\left\mid 2 x + 3 \right\mid\right) + C$

#### Explanation:

Note that

$6 {x}^{2} + 5 x - 6 = 6 {x}^{2} + 9 x - 4 x - 6$

$= 3 x \left(2 x + 3\right) - 2 \left(2 x + 3\right)$

$= \left(3 x - 2\right) \left(2 x + 3\right)$

The partial fraction decomposition can then be set up as:

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

Multiplying both sides by $\left(3 x - 2\right) \left(2 x + 3\right)$, we see that

$13 x = A \left(2 x + 3\right) + B \left(3 x - 2\right)$

Letting $x = - \frac{3}{2}$:

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

$- \frac{39}{2} = A \left(- 3 + 3\right) + B \left(- \frac{9}{2} - 2\right)$

$- \frac{39}{2} = B \left(- \frac{13}{2}\right)$

$3 = B$

Letting $x = \frac{2}{3}$:

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

$\frac{26}{3} = A \left(\frac{4}{3} + 3\right) + B \left(2 - 2\right)$

$\frac{26}{3} = A \left(\frac{13}{3}\right)$

$2 = A$

Thus,

$\frac{13 x}{\left(3 x - 2\right) \left(2 x + 3\right)} = \frac{2}{3 x - 2} + \frac{3}{2 x + 3}$

So,

$\int \frac{13 x}{6 {x}^{2} + 5 x - 6} \mathrm{dx} = 2 \int \frac{1}{3 x - 2} \mathrm{dx} + 3 \int \frac{1}{2 x + 3} \mathrm{dx}$

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

$= \frac{2}{3} \ln \left(\left\mid 3 x - 2 \right\mid\right) + \frac{3}{2} \ln \left(\left\mid 2 x + 3 \right\mid\right) + C$