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

Nov 21, 2016

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

#### Explanation:

Let us find the partial fraction decomposition of the integrand:

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

This is an identity and valid $\forall x \in \mathbb{R}$
Put $x = - 2 \implies - 6 = - 3 A + 0 \implies A = 2$
Put $x = 1 \implies 3 = 0 + B \left(3\right) \implies B = 1$

So the partial fraction decomposition is:

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

And so the integral can be written as:

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