# How do you integrate (2x) / (4x^2 + 12x + 9) using partial fractions?

Dec 28, 2016

$= \frac{1}{2} \ln | 2 x + 3 | + \frac{3}{2 \left(2 x + 3\right)} + C$.
You don't use partial fractions, because the denominator is a perfect square.

#### Explanation:

$\int \frac{2 x}{4 {x}^{2} + 12 x + 9} \mathrm{dx}$
$= \int \frac{2 x}{2 x + 3} ^ 2 \mathrm{dx}$
So substitute $u = 2 x + 3$, $\mathrm{dx} = \frac{\mathrm{du}}{2}$:
$= \frac{1}{2} \int \frac{u - 3}{u} ^ 2 \mathrm{du}$
$= \frac{1}{2} \int {u}^{-} 1 - 3 {u}^{-} 2 \mathrm{dx}$
$= \frac{1}{2} \ln | 2 x + 3 | + \frac{3}{2 \left(2 x + 3\right)} + C$