# How do you integrate int (x)/((x^2-4)(x-3) dx using partial fractions?

Feb 1, 2017

$\int \frac{x}{\left({x}^{2} - 4\right) \left(x - 3\right)} \setminus \mathrm{dx} = - \frac{1}{2} \ln | x - 2 | - \frac{1}{10} \ln | x + 2 | + \frac{3}{5} \ln | x - 3 | + C$

#### Explanation:

We want to evaluate the integral:

$\int \frac{x}{\left({x}^{2} - 4\right) \left(x - 3\right)} \setminus \mathrm{dx}$

We can factorise the integrand:

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

And so the integrand will have a partial fraction decomposition of the form:

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

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

Subs $x = 2 \setminus \setminus \setminus \setminus \setminus \implies 2 \setminus \setminus \setminus \setminus \setminus = A \left(4\right) \left(- 1\right) \setminus \setminus \setminus \setminus \setminus \implies A = - \frac{1}{2}$
Subs $x = - 2 \implies - 2 = B \left(- 4\right) \left(- 5\right) \implies B = - \frac{1}{10}$
Subs $x = 3 \setminus \setminus \setminus \setminus \setminus \implies 3 \setminus \setminus \setminus \setminus \setminus = C \left(1\right) \left(5\right) \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies C = \frac{3}{5}$

So the partial fraction decomposition of the integrand is:

$\frac{x}{\left(x - 2\right) \left(x + 2\right) \left(x - 3\right)} \equiv \frac{- \frac{1}{2}}{x - 2} + \frac{- \frac{1}{10}}{x + 2} + \frac{\frac{3}{5}}{x - 3}$

So the integral becomes:

$\int \frac{x}{\left({x}^{2} - 4\right) \left(x - 3\right)} \setminus \mathrm{dx} = \int \setminus \frac{- \frac{1}{2}}{x - 2} - \frac{\frac{1}{10}}{x + 2} + \frac{\frac{3}{5}}{x - 3} \setminus \mathrm{dx}$
$\text{ } = - \frac{1}{2} \ln | x - 2 | - \frac{1}{10} \ln | x + 2 | + \frac{3}{5} \ln | x - 3 | + C$