# Compute int_0^2 1/(x-1)dx

Compute ${\int }_{0}^{2}\frac{1}{x-1}dx$
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, which lies between the upper and lower integration bounds.
To determine if the integral is convergent, you must then split the integral at x=1, and then use limits to evaluate:
${\int }_{0}^{2}\frac{1}{x-1}dx$
$={\int }_{0}^{1}\frac{1}{x-1}dx+{\int }_{1}^{2}\frac{1}{x-1}dx$ (Rewrite as a sum)
$\underset{m\to {1}^{-}}{lim}{\int }_{0}^{m}\frac{1}{x-1}dx+\underset{m\to {1}^{-}}{lim}{\int }_{n}^{2}\frac{1}{x-1}dx$ (Rewrite using limits)
$\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|x-1|\right)\left({\mid }_{0}^{m}\right)+\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|x-1|\right){\mid }_{n}^{2}$ (Integrate)
$\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|m-1|-\mathrm{ln}|0-1|\right)+\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|2-1|-\mathrm{ln}|n-1|\right)$ (Evaluate)
$\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|m-1|-\mathrm{ln}1\right)+\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}1-\mathrm{ln}|n-1|\right)$ (Simplify)
$\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|m-1|\right)+\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|n-1|\right)$ (Simplify using $\mathrm{ln}1=0$)
$\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|m-1|\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{m\to {1}^{-}}{lim}\left(\mathrm{ln}|n-1|\right)$ aren't finite since $\mathrm{ln}0$ is not defined.Therefore, the integral is divergent.