Definite integral of an absolute value with "undefined result"

Question

I am trying to find a correct result of definite integral $\int_0^2\dfrac{x^2-x+1}{x-1} dx$ .

But I got after integration $\left[\dfrac{x^2}{2}+\ln(x-1)\right]_0^2$ and $\ln$ is undefined in $\mathbb{R}$ for negative numbers.

I tried wolfram alpha, but I would like to see the answer step by step which unavailable for this example wolfram

Please could anyone tell what should be the correct result.

Answer 1

The numerator of $f(x)$ is clearly positive on the interval $[0,2]$, so $f < 0$ on $[0,1)$ and $f > 0$ on $(1,2]$. $f$ is undefined at $x = 1$. The given integral therefore is improper, and converges in the Riemann sense if and only if each of $$\lim_{\epsilon \to 0^+} \int_{x=0}^{1-\epsilon} f(x) \, dx$$ and $$\lim_{\epsilon \to 0^+} \int_{x=1+\epsilon}^2 f(x) \, dx$$ exist and are finite. Since they are not, the integral does not converge.

Answer 2

Note that $\frac{x^2-x+1}{x-1} = x+\frac{1}{x-1}$. The first part of the sum is integrable on $[0,2]$, the latter is not.