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

Oct 2, 2016

$\int \frac{x}{{x}^{3} - {x}^{2} - 6 x} \mathrm{dx} = \frac{1}{5} \ln \left\mid x - 3 \right\mid - \frac{1}{5} \ln \left\mid x + 2 \right\mid + C$

#### Explanation:

Quick version

Note that both the numerator and denominator are divisible by $x$, so we find:

$\frac{x}{{x}^{3} - {x}^{2} - 6 x} = \frac{1}{{x}^{2} - x - 6} = \frac{1}{\left(x - 3\right) \left(x + 2\right)} = \frac{1}{5 \left(x - 3\right)} - \frac{1}{5 \left(x + 2\right)}$

So:

$\int \frac{x}{{x}^{3} - {x}^{2} - 6 x} \mathrm{dx} = \int \frac{1}{5 \left(x - 3\right)} - \frac{1}{5 \left(x + 2\right)} \mathrm{dx}$

$\textcolor{w h i t e}{\int \frac{x}{{x}^{3} - {x}^{2} - 6 x} \mathrm{dx}} = \frac{1}{5} \ln \left\mid x - 3 \right\mid - \frac{1}{5} \ln \left\mid x + 2 \right\mid + C$

Notes

$\frac{1}{\left(x - 3\right) \left(x + 2\right)} = \frac{A}{x - 3} + \frac{B}{x + 2}$

$\textcolor{w h i t e}{\frac{1}{\left(x - 3\right) \left(x + 2\right)}} = \frac{A \left(x + 2\right) + B \left(x - 3\right)}{\left(x - 3\right) \left(x + 2\right)}$

$\textcolor{w h i t e}{\frac{1}{\left(x - 3\right) \left(x + 2\right)}} = \frac{\left(A + B\right) x + \left(2 A - 3 B\right)}{\left(x - 3\right) \left(x + 2\right)}$

Equating coefficients, we have:

$\left\{\begin{matrix}A + B = 0 \\ 2 A - 3 B = 1\end{matrix}\right.$

Adding $3$ times the first equation to the second, we get:

$5 A = 1$

Hence $A = \frac{1}{5}$ and $B = - \frac{1}{5}$

So:

$\frac{1}{\left(x - 3\right) \left(x + 2\right)} = \frac{1}{5 \left(x - 3\right)} - \frac{1}{5 \left(x + 2\right)}$

$\textcolor{w h i t e}{}$
Alternatively, we could find $A$ and $B$ using Heaviside's cover up method:

$A = \frac{1}{\textcolor{b l u e}{3} + 2} = \frac{1}{5}$

$B = \frac{1}{\textcolor{b l u e}{- 2} - 3} = \frac{1}{- 5} = - \frac{1}{5}$

$\textcolor{w h i t e}{}$
The method I actually used was:

If $\frac{1}{\left(x - 3\right) \left(x + 2\right)} = \frac{A}{x - 3} + \frac{B}{x + 2}$

then when we express in terms of a common denominator we will need $A$ and $B$ to be the same size but opposite signs in order that the $x$ term cancels out.

If try $\frac{1}{x - 3} - \frac{1}{x + 2}$ then we get $2 - \left(- 3\right) = 2 + 3 = 5$, so that's $5$ times too large.

Hence $A = \frac{1}{5}$ and $B = - \frac{1}{5}$