# How do you integrate (x+2)/(x^2+8x+16) using partial fractions?

##### 1 Answer
Dec 25, 2016

$\int \frac{x + 2}{{x}^{2} + 8 x + 16} \mathrm{dx} = \ln | x + 4 | + \frac{2}{x + 4} + C$

#### Explanation:

First you factorize the denominator of the rational function:

${x}^{2} + 8 x + 16 = {\left(x + 4\right)}^{2}$

Now you can write:

$\frac{x + 2}{{x}^{2} + 8 x + 16} = \frac{x + 2}{{\left(x + 4\right)}^{2}} = \frac{x + 4 - 2}{{\left(x + 4\right)}^{2}} = \frac{1}{x + 4} - \frac{2}{{\left(x + 4\right)}^{2}}$

The integral is then:

$\int \frac{x + 2}{{x}^{2} + 8 x + 16} \mathrm{dx} = \int \frac{\mathrm{dx}}{x + 4} - 2 \int \frac{\mathrm{dx}}{{\left(x + 4\right)}^{2}} = \ln | x + 4 | + \frac{2}{x + 4} + C$