# How do you integrate (9x)/(9x^2+3x-2) using partial fractions?

Nov 22, 2016

The answer is $= = \frac{1}{3} \ln \left(3 x - 1\right) + \frac{2}{3} \ln \left(3 x + 2\right) + C$

#### Explanation:

Let's factorise the denominator

$9 {x}^{2} + 3 x - 2 = \left(3 x - 1\right) \left(3 x + 2\right)$

Now, we can start by the decomposition into partial fractions

$\frac{9 x}{9 {x}^{2} + 3 x - 2} = \frac{9 x}{\left(3 x - 1\right) \left(3 x + 2\right)} = \frac{A}{3 x - 1} + \frac{B}{3 x + 2}$

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

Therefore,

$9 x = A \left(3 x + 2\right) + B \left(3 x - 1\right)$

Let $x = 0$, $\implies$, $0 = 2 A - B$

Coefficients of $x$,

$9 = 3 A + 3 B$, $\implies$, $A + B = 3$

Solving for A and B, we get $A = 1$ and $B = 2$

So,
$\frac{9 x}{9 {x}^{2} + 3 x - 2} = \frac{1}{3 x - 1} + \frac{2}{3 x + 2}$

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

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