# How do you integrate int(x)/((x+4)(x+2)(2x-11)) using partial fractions?

Jul 28, 2016

$\int \frac{x}{\left(x + 4\right) \left(x + 2\right) \left(2 x - 11\right)} \mathrm{dx}$

$= - \frac{2}{19} \ln \left(\left\mid x + 4 \right\mid\right) + \frac{1}{15} \ln \left(\left\mid x + 2 \right\mid\right) + \frac{11}{285} \ln \left(\left\mid 2 x - 11 \right\mid\right) + C$

#### Explanation:

$\frac{x}{\left(x + 4\right) \left(x + 2\right) \left(2 x - 11\right)} = \frac{A}{x + 4} + \frac{B}{x + 2} + \frac{C}{2 x - 11}$

Use Heaviside's cover up method to find:

$A = \frac{- 4}{\left(\left(- 4\right) + 2\right) \left(2 \left(- 4\right) - 11\right)} = \frac{- 4}{\left(- 2\right) \left(- 19\right)} = - \frac{2}{19}$

$B = \frac{- 2}{\left(\left(- 2\right) + 4\right) \left(2 \left(- 2\right) - 11\right)} = \frac{- 2}{\left(2\right) \left(- 15\right)} = \frac{1}{15}$

$C = \frac{\frac{11}{2}}{\left(\left(\frac{11}{2}\right) + 4\right) \left(\left(\frac{11}{2}\right) + 2\right)} = \frac{22}{\left(11 + 8\right) \left(11 + 4\right)} = \frac{22}{285}$

So:

$\int \frac{x}{\left(x + 4\right) \left(x + 2\right) \left(2 x - 11\right)} \mathrm{dx}$

$= \int \left(- \frac{2}{19 \left(x + 4\right)} + \frac{1}{15 \left(x + 2\right)} + \frac{22}{285 \left(2 x - 11\right)}\right) \mathrm{dx}$

$= - \frac{2}{19} \ln \left(\left\mid x + 4 \right\mid\right) + \frac{1}{15} \ln \left(\left\mid x + 2 \right\mid\right) + \frac{11}{285} \ln \left(\left\mid 2 x - 11 \right\mid\right) + C$

Notice the coefficient $\frac{11}{285}$, not $\frac{22}{285}$. When differentiated, the $2 x$ results in a factor $2$.