# \int(x^3+x)/(x-1)dx?

## Check answers? Long division got me $\setminus \int {x}^{2} + x + 2 + \frac{2}{x - 1} \mathrm{dx}$ Solving: $\setminus \int \frac{2}{x - 1} \mathrm{dx}$ by u-sub ($u = x - 1$ and $\mathrm{du} = \mathrm{dx}$) becomes $2 \setminus \int \frac{1}{u} \mathrm{du} = 2 \setminus \ln u + C = 2 \setminus \ln | x - 1 | + C$ $\setminus \therefore \setminus \int {x}^{2} + x + 2 + \frac{2}{x - 1} \mathrm{dx} = {x}^{3} / 3 + {x}^{2} / 2 + 2 x + 2 \setminus \ln | x - 1 | + C$

Apr 2, 2018

$I = {x}^{3} / 3 + {x}^{2} / 2 + 2 x + 2 \ln | x - 1 | + c$

#### Explanation:

Here,

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

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

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

$= {x}^{3} / 3 + {x}^{2} / 2 + x + x + 2 \ln | x - 1 | + c$

$= {x}^{3} / 3 + {x}^{2} / 2 + 2 x + 2 \ln | x - 1 | + c$