# How do you integrate int (5x+12)/(x^2+5x+6) using partial fractions?

Feb 7, 2016

$\ln \left(x + 3\right) + \ln \left(x + 2\right) + C$

#### Explanation:

To begin we must take:

$\frac{5 x + 12}{{x}^{2} + 5 x + 6}$ and decompose it into its partial fractions. First factorise the denominator and then split the fraction up as follows:

$\frac{5 x + 12}{\left(x + 3\right) \left(x + 2\right)} = \frac{A}{x + 3} + \frac{B}{x + 2}$

Now, if we multiply the whole thing through by $\left(x + 3\right) \left(x + 2\right)$ then we should get an equation that will allow us to solve for $A$ and $B$.

$\to 5 x + 12 = A \left(x + 2\right) + B \left(x + 3\right)$

Now, to find $A$ set $x = 3$ to cancel the second term and we get:

$5 \left(- 3\right) + 12 = A \left(- 3 + 2\right) + B \left(- 3 + 3\right)$
$- 3 = - A \to A = 3$

Now set $x = - 2$ to obtain the value for for $B$.

$\to 5 \left(- 2\right) + 12 = A \left(- 2 + 2\right) + B \left(- 2 + 3\right) \to B = 2$

So now we have that $A = 3$ and $B = 2$ we can re write the fraction given in the question as:

$\frac{5 x + 12}{{x}^{2} + 5 x + 6} = \frac{3}{x + 3} + \frac{2}{x + 2}$

So we can now integrate:

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