# How do you integrate int (3x)/((x + 2)(x - 1)) using partial fractions?

Dec 10, 2016

$\int \frac{3 x}{\left(x + 2\right) \left(x - 1\right)} = \ln \left(A {\left(x + 2\right)}^{2} | x - 1 |\right)$

#### Explanation:

The partial fraction decomposition of the integrand will be of the form:

$\frac{3 x}{\left(x + 2\right) \left(x - 1\right)} = \frac{A}{x + 2} + \frac{B}{x - 1}$
$\text{ } = \frac{A \left(x - 1\right) + B \left(x + 2\right)}{\left(x + 2\right) \left(x - 1\right)}$
$\therefore 3 x \text{ } = A \left(x - 1\right) + B \left(x + 2\right)$

Put $x = - 2 \implies - 6 = A \left(- 3\right) \implies A = 2$
Put $x = 1 \text{ "=> 3=B(3) " } \implies B = 1$

Hence,

$\frac{3 x}{\left(x + 2\right) \left(x - 1\right)} = \frac{2}{x + 2} + \frac{1}{x - 1}$

And so:

$\int \frac{3 x}{\left(x + 2\right) \left(x - 1\right)} = \int \left(\frac{2}{x + 2} + \frac{1}{x - 1}\right) \mathrm{dx}$
$\text{ } = 2 \int \frac{1}{x + 2} \mathrm{dx} + \int \frac{1}{x - 1} \mathrm{dx}$
$\text{ } = 2 \ln | x + 2 | + \ln | x - 1 | + c$
$\text{ } = \ln {\left(| x + 2 |\right)}^{2} + \ln | x - 1 | + \ln A$ ($c = \ln A$)
$\text{ } = \ln \left(A {\left(x + 2\right)}^{2} | x - 1 |\right)$