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

Oct 18, 2016

$= {x}^{2} / 2 + x - 2 \ln x + 3 \ln \left(x - 1\right) + C$

#### Explanation:

First make a long division to determine
$\frac{{x}^{3} + 2}{{x}^{2} - x} = x + 1 + \frac{x + 2}{{x}^{2} - x}$
To simplify the fraction we use partial fraction
$\frac{x + 2}{{x}^{2} - x} = \frac{x + 2}{x \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x - 1} = \frac{A \left(x - 1\right) + B x}{x \left(x - 1\right)}$
So we determine A and B
$x + 2 = A \left(x - 1\right) + B x$
Cmparing the coefficients, we find
$1 = A + B$ and $2 = - A$ and we conclude $B = 3$
$\int \frac{\left({x}^{3} + 2\right) \mathrm{dx}}{{x}^{2} - x} = \int \left(x + 1\right) \mathrm{dx} - \int 2 \frac{\mathrm{dx}}{x} + \int 3 \frac{\mathrm{dx}}{x - 1}$
$= {x}^{2} / 2 + x - 2 \ln x + 3 \ln \left(x - 1\right) + C$