Find the integral of 4x/ (x^2 - 4 )(x-3) dx ?

Aug 14, 2018

$I = - \frac{2}{5} \left\{5 \ln | x - 2 | + \ln | x + 2 | - 6 \ln | x - 3 |\right\} + c$

Explanation:

Here,
$I = \int \frac{4 x}{\left({x}^{2} - 4\right) \left(x - 3\right)} \mathrm{dx} = \int \frac{4 x}{\left(x - 2\right) \left(x + 2\right) \left(x - 3\right)} \mathrm{dx}$

To obtain partial fractions :

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

$4 x = A \left(x + 2\right) \left(x - 3\right) + B \left(x - 2\right) \left(x - 3\right) + C \left(x - 2\right) \left(x + 2\right)$

$\therefore x = 2 \implies 8 = A \left(4\right) \left(- 1\right) \implies A = - 2$

$x = - 2 \implies - 8 = B \left(- 4\right) \left(- 5\right) \implies B = - \frac{2}{5}$

$x = 3 \implies 12 = C \left(1\right) \left(5\right) \implies C = \frac{12}{5}$

$\therefore I = \int \left[\frac{- 2}{x - 2} + \frac{- \frac{2}{5}}{x + 2} + \frac{\frac{12}{5}}{x - 3}\right] \mathrm{dx}$

$\therefore I = - \frac{2}{5} \int \left[\frac{5}{x - 2} + \frac{1}{x + 2} - \frac{6}{x - 3}\right] \mathrm{dx}$

$\therefore I = - \frac{2}{5} \left\{5 \ln | \left(x - 2\right) | + \ln | \left(x + 2\right) | - 6 \ln | \left(x - 3\right) |\right\} + c$