How do you integrate f(x)=(x)/((x+4)(x-2)(2x-2)) using partial fractions?

May 29, 2017

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

$= - \frac{1}{15} \ln \left\mid x + 4 \right\mid + \frac{1}{6} \ln \left\mid x - 2 \right\mid - \frac{1}{10} \ln \left\mid x - 1 \right\mid + C$

Explanation:

$\frac{x}{\left(x + 4\right) \left(x - 2\right) \left(2 x - 2\right)} = \frac{x}{2 \left(x + 4\right) \left(x - 2\right) \left(x - 1\right)}$

$\textcolor{w h i t e}{\frac{x}{\left(x + 4\right) \left(x - 2\right) \left(2 x - 2\right)}} = \frac{A}{x + 4} + \frac{B}{x - 2} + \frac{C}{x - 1}$

Use Oliver Heaviside's cover up method to find $A$, $B$ and $C$:

$A = \frac{\textcolor{b l u e}{- 4}}{2 \left(\left(\textcolor{b l u e}{- 4}\right) - 2\right) \left(\left(\textcolor{b l u e}{- 4}\right) - 1\right)} = \frac{- 4}{2 \left(- 6\right) \left(- 5\right)} = - \frac{1}{15}$

$B = \frac{\textcolor{b l u e}{2}}{2 \left(\textcolor{b l u e}{2} + 4\right) \left(\textcolor{b l u e}{2} - 1\right)} = \frac{2}{2 \left(6\right) \left(1\right)} = \frac{1}{6}$

$C = \frac{\textcolor{b l u e}{1}}{2 \left(\textcolor{b l u e}{1} + 4\right) \left(\textcolor{b l u e}{1} - 2\right)} = \frac{1}{2 \left(5\right) \left(- 1\right)} = - \frac{1}{10}$

So:

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

$= \int \left(- \frac{1}{15} \left(\frac{1}{x + 4}\right) + \frac{1}{6} \left(\frac{1}{x - 2}\right) - \frac{1}{10} \left(\frac{1}{x - 1}\right)\right) \mathrm{dx}$

$= - \frac{1}{15} \ln \left\mid x + 4 \right\mid + \frac{1}{6} \ln \left\mid x - 2 \right\mid - \frac{1}{10} \ln \left\mid x - 1 \right\mid + C$