# How do you integrate (5x)/(2x^2+11x+12) using partial fractions?

Nov 24, 2016

$\int \left(\frac{5 x}{2 {x}^{2} + 11 x + 12}\right) \mathrm{dx} = 4 \ln | x + 4 | - \frac{3}{2} \ln | 2 x + 3 | + C$

#### Explanation:

We start by factoring the denominator.

$2 {x}^{2} + 11 x + 12 = 2 {x}^{2} + 8 x + 3 x + 12 = 2 x \left(x + 4\right) + 3 \left(x + 4\right) = \left(2 x + 3\right) \left(x + 4\right)$.

The partial fraction decomposition will therefore be of the form:

A/(2x + 3) + B/(x + 4) = (5x)/((2x + 3)(x + 4)

$A \left(x + 4\right) + B \left(2 x + 3\right) = 5 x$

$A x + 4 A + 2 B x + 3 B = 5 x$

$\left(A + 2 B\right) x + \left(4 A + 3 B\right) = 5 x$

We now write a systems of equations.

$\left\{\begin{matrix}A + 2 B = 5 \\ 4 A + 3 B = 0\end{matrix}\right.$

Solve:

$A = 5 - 2 B$

$4 \left(5 - 2 B\right) + 3 B = 0$

$20 - 8 B + 3 B = 0$

$- 5 B = - 20$

$B = 4$

$A + 2 \left(4\right) = 5$

$A = - 3$

Hence, the partial fraction decomposition is $\frac{4}{x + 4} - \frac{3}{2 x + 3}$. The integral becomes

$\int \left(\frac{4}{x + 4} - \frac{3}{2 x + 3}\right) \mathrm{dx}$

We know that $\int \left(\frac{1}{u}\right) \mathrm{du} = \ln | u | + C$. Therefore:

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

Hopefully this helps!