# How do you use partial fraction decomposition to decompose the fraction to integrate (-x - 38)/(2x^2 + 9x - 5) ?

Nov 29, 2016

$\int \frac{- x - 38}{2 {x}^{2} + 9 x - 5} \mathrm{dx} = 3 \ln | x + 5 | - \frac{7}{2} \ln | 2 x - 1 | + c$

#### Explanation:

First we need to factorise the denominator.

To do this we need to find factors of $- 10$ (coefficient of ${x}^{2} \times$ coefficient of constant$= 2 \times - 5$) that add up tp 9 (the coefficient of $x = 9$). So the factors we seek are $- 1$ and $10$

$\therefore 2 {x}^{2} + 9 x - 5 = 2 {x}^{2} + 10 x - x - 5$
$\therefore 2 {x}^{2} + 9 x - 5 = 2 x \left(x + 5\right) - \left(x + 5\right)$
$\therefore 2 {x}^{2} + 9 x - 5 = \left(x + 5\right) \left(2 x - 1\right)$

Therefore we can write the integrand as follows:

$\frac{- x - 38}{2 {x}^{2} + 9 x - 5} = \frac{- x - 38}{\left(x + 5\right) \left(2 x - 1\right)}$

And thge partial fraction decomposition will be:

$\setminus \setminus \setminus \setminus \setminus \frac{- x - 38}{2 {x}^{2} + 9 x - 5} = \frac{A}{x + 5} + \frac{B}{2 x - 1}$
$\therefore \frac{- x - 38}{2 {x}^{2} + 9 x - 5} = \frac{A \left(2 x - 1\right) + B \left(x + 5\right)}{\left(x + 5\right) \left(2 x - 1\right)}$
$\therefore \setminus \setminus \setminus \setminus \left(- x - 38\right) = A \left(2 x - 1\right) + B \left(x + 5\right)$

Subs $x = - 5 \implies - \left(- 5\right) - 38 = A \left(- 10 - 1\right) + 0$
$\therefore - 11 A = - 33 \implies A = 3$

And Sub $x = \frac{1}{2} \implies - \frac{1}{2} - 38 = B \left(\frac{1}{2} + 5\right)$
$\therefore \frac{11}{2} B = - \frac{77}{2} \implies B = - 7$

Hence, the partial fraction decomposition of the integrand is:

$\frac{- x - 38}{2 {x}^{2} + 9 x - 5} = \frac{3}{x + 5} - \frac{7}{2 x - 1}$

And so;

$\int \frac{- x - 38}{2 {x}^{2} + 9 x - 5} \mathrm{dx} = \int \frac{3}{x + 5} - \frac{7}{2 x - 1} \mathrm{dx}$

And integrating we get:

$\int \frac{- x - 38}{2 {x}^{2} + 9 x - 5} \mathrm{dx} = 3 \ln | x + 5 | - \frac{7}{2} \ln | 2 x - 1 | + c$