# How do you use partial fraction decomposition to decompose the fraction to integrate 11/(3x^2-7x-6)?

Apr 4, 2018

$\int \frac{11}{3 {x}^{2} - 7 x - 6} \mathrm{dx} = \ln \left\mid \frac{x - 3}{3 x + 2} \right\mid + C$

#### Explanation:

Factorize the denominator by solving the equation:

$3 {x}^{2} - 7 x - 6 = 0$

$x = \frac{7 \pm \sqrt{49 + 72}}{6} = \frac{7 \pm \sqrt{121}}{6} = \frac{7 \pm 11}{6} = \left\{\begin{matrix}- \frac{2}{3} \\ 3\end{matrix}\right.$

Then:

$3 {x}^{2} - 7 x - 6 = 3 \left(x + \frac{2}{3}\right) \left(x - 3\right) = \left(3 x + 2\right) \left(x - 3\right)$

Apply partial fractions decomposition:

$\frac{11}{3 {x}^{2} - 7 x - 6} = \frac{A}{3 x + 2} + \frac{B}{x - 3}$

$\frac{11}{3 {x}^{2} - 7 x - 6} = \frac{A \left(x - 3\right) + B \left(3 x + 2\right)}{3 {x}^{2} - 7 x - 6}$

$11 = \left(A + 3 B\right) x - 3 A + 2 B$

$\left\{\begin{matrix}A + 3 B = 0 \\ - 3 A + 2 B = 11\end{matrix}\right.$

$\left\{\begin{matrix}A = - 3 \\ B = 1\end{matrix}\right.$

So:

$\frac{11}{3 {x}^{2} - 7 x - 6} = - \frac{3}{3 x + 2} + \frac{1}{x - 3}$

$\int \frac{11}{3 {x}^{2} - 7 x - 6} \mathrm{dx} = - \int \frac{3 \mathrm{dx}}{3 x + 2} + \int \frac{\mathrm{dx}}{x - 3}$

$\int \frac{11}{3 {x}^{2} - 7 x - 6} \mathrm{dx} = - \int \frac{d \left(3 x + 2\right)}{3 x + 2} + \int \frac{d \left(x - 3\right)}{x - 3}$

$\int \frac{11}{3 {x}^{2} - 7 x - 6} \mathrm{dx} = - \ln \left\mid 3 x + 2 \right\mid + \ln \left\mid x - 3 \right\mid + C$

$\int \frac{11}{3 {x}^{2} - 7 x - 6} \mathrm{dx} = \ln \left\mid \frac{x - 3}{3 x + 2} \right\mid + C$