Find the following integral: $\int \frac{x^{3}-x^{2}+x-1}{x-1} \, dx$

To evaluate the integral $\int \frac{x^{3}-x^{2}+x-1}{x-1} \, dx$,we first simplify the integrand by factorising the numerator.
$\frac{x^{3}-x^{2}+x-1}{x-1} = \frac{x^{2}(x-1) + 1(x-1)}{x-1} = \frac{(x^{2}+1)(x-1)}{x-1}$
For $x \neq 1$,this simplifies to $x^{2}+1$.
Now,we integrate the simplified expression:
$\int (x^{2}+1) \, dx = \int x^{2} \, dx + \int 1 \, dx$
Using the power rule $\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C$,we get:
$= \frac{x^{3}}{3} + x + C$
where $C$ is an arbitrary constant.