# How do you integrate (x^2+2x-1)/(x^3-x)dx?

Oct 13, 2016

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

#### Explanation:

$\frac{{x}^{2} + 2 x - 1}{{x}^{3} - x} = \frac{{x}^{2} + 2 x - 1}{x \left(x - 1\right) \left(x + 1\right)} = \frac{a}{x} + \frac{b}{x - 1} + \frac{c}{x + 1}$

Use Heaviside's cover up method to find $a , b , c$:

$a = \frac{{\left(\textcolor{b l u e}{0}\right)}^{2} + 2 \left(\textcolor{b l u e}{0}\right) - 1}{\left(\left(\textcolor{b l u e}{0}\right) - 1\right) \left(\left(\textcolor{b l u e}{0}\right) + 1\right)} = \frac{- 1}{\left(- 1\right) \left(1\right)} = 1$

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

$c = \frac{{\left(\textcolor{b l u e}{- 1}\right)}^{2} + 2 \left(\textcolor{b l u e}{- 1}\right) - 1}{\left(\textcolor{b l u e}{- 1}\right) \left(\left(\textcolor{b l u e}{- 1}\right) - 1\right)} = \frac{- 2}{\left(- 1\right) \left(- 2\right)} = - 1$

So:

$\int \frac{{x}^{2} + 2 x - 1}{{x}^{3} - x} \mathrm{dx} = \int \frac{1}{x} + \frac{1}{x - 1} - \frac{1}{x + 1} \mathrm{dx}$

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