# How do you use partial fraction decomposition to decompose the fraction to integrate 26/(6x^2+5x-6)?

Jun 8, 2018

See process below

#### Explanation:

First of all, we factorize the polynomial $6 {x}^{2} + 5 x - 6$

$6 {x}^{2} + 5 x - 6 = 0$

$x = \frac{- 5 \pm \sqrt{25 + 144}}{12} = \frac{- 5 \pm 13}{6}$ this give two solutions

${x}_{1} = - 3$ and ${x}_{2} = \frac{4}{3}$

Thus we have $6 {x}^{2} + 5 x - 6 = \left(x + 3\right) \left(x - \frac{4}{3}\right)$

$\frac{1}{6 {x}^{2} + 5 x - 6} = \frac{A}{x + 3} + \frac{B}{x - \frac{4}{3}}$ transposing terms and equalizing we have

$A + B = 0$
$3 B - \frac{4}{3} A = 1$ from here we find $A = - \frac{3}{13}$ and $B = \frac{3}{13}$

Then finally we have $\frac{1}{6 {x}^{2} + 5 x - 6} = \frac{- \frac{3}{13}}{x + 3} + \frac{\frac{3}{13}}{x - \frac{4}{3}}$