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

Jun 20, 2016

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

#### Explanation:

As $\frac{{x}^{2} + 2 x - 1}{{x}^{3} - x} = \frac{{x}^{2} + 2 x - 1}{x \left(x + 1\right) \left(x - 1\right)}$, let us convert them into partial fractions

$\frac{{x}^{2} + 2 x - 1}{{x}^{3} - x} \Leftrightarrow \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}$ or

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

$\frac{{x}^{2} + 2 x - 1}{{x}^{3} - x} = \frac{\left(A + B + C\right) {x}^{2} + \left(- B + C\right) x - A}{x \left(x + 1\right) \left(x - 1\right)}$ i.e.

$A + B + C = 1$, $- B + C = 2$ and $A = 1$,

which gives $C = 1$ and $B = - 1$ and hence

$\frac{{x}^{2} + 2 x - 1}{{x}^{3} - x} = \frac{1}{x} - \frac{1}{x + 1} + \frac{1}{x - 1}$ and

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

= $\ln x - \ln \left(x + 1\right) + \ln \left(x - 1\right) + c$