# How do you integrate (3x+11)/(x^2-x-6) using partial fractions?

Dec 19, 2016

The answer is $4 \ln | x - 3 | - \ln | x + 2 | + C$

#### Explanation:

The denominator can be factored as $\left(x - 3\right) \left(x + 2\right)$, so:

$\frac{A}{x - 3} + \frac{B}{x + 2} = \frac{3 x + 11}{\left(x - 3\right) \left(x + 2\right)}$

$A \left(x + 2\right) + B \left(x - 3\right) = 3 x + 11$

$A x + 2 A + B x - 3 B = 3 x + 11$

$\left(A + B\right) x + \left(2 A - 3 B\right) = 3 x + 11$

We now write a systems of equations to solve for $A$ and $B$.

$\left\{\begin{matrix}A + B = 3 \\ 2 A - 3 B = 11\end{matrix}\right.$

Solving:

$B = 3 - A \to 2 A - 3 \left(3 - A\right) = 11 \to 2 A - 9 + 3 A = 11 \to 5 A = 20 \to A = 4$

$A + B = 3 \to 4 + B = 3 \to B = - 1$

So, $A = 4$ and $B = - 1$.

The partial fraction decomposition is therefore $\textcolor{b l u e}{\frac{4}{x - 3} - \frac{1}{x + 2}}$.

This can be integrated using the rule $\int \frac{1}{x} \mathrm{dx} = \ln | x | + C$.

$\int \left(\frac{4}{x - 3} - \frac{1}{x + 2}\right) = 4 \ln | x - 3 | - \ln | x + 2 | + C$

Hopefully this helps!