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

Dec 15, 2016

$\int \frac{\mathrm{dx}}{{x}^{2} + 3 x + 2} = \ln | \frac{x + 1}{x + 2} | + C$

#### Explanation:

First you need to factorize the denominator:

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

Now put:

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

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

As the denominators are the same, so must be the numerators:

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

and equating the coefficients of the same degree in $x$:

$A + B = 0$
$2 A + B = 1$

Solving this system we get:

$A = 1$, $B = - 1$

so that:

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