Calculating integrals with asymptotes? Find $\displaystyle\int^2_0 \dfrac{1}{(1-x)^2} dx$.

Is there a way of doing this without considering the asymptote at $x=1$? What if you didn't know at first that there was indeed an asymptote at this point?
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One alternative is the Cauchy Principal Value $\ds{\color{#f00}{\mrm{P.V.}}}$:
\begin{align}
\color{#f00}{\mrm{P.V.}\int_{0}^{2}{\dd x \over \pars{x - 1}^{2}}} &amp;
\,\,\,\stackrel{\mbox{def.}}{=}\,\,\,
\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{0}^{1 - \epsilon}{\dd x \over \pars{x - 1}^{2}} +
\int_{1 + \epsilon}^{2}{\dd x \over \pars{x - 1}^{2}}}
\\[5mm] &amp; =
\lim_{\epsilon \to 0^{+}}\bracks{%
{1 \over 1 - \pars{1 -\epsilon}} - {1 \over 1 - 0} + {1 \over 1 -2} -
{1 \over 1 - \pars{1 + \epsilon}}} =
\lim_{\epsilon \to 0^{+}}\pars{{2 \over \epsilon} - 2}
\\[5mm] &amp; = \color{#f00}{\infty}
\end{align}


  Even the Cauchy Principal Value $\underline{diverges}$.
I upvoted the other answer because I agree. However there are a few Nice identities

https://en.m.wikipedia.org/wiki/Sokhotski%E2%80%93Plemelj_theorem

https://en.m.wikipedia.org/wiki/Hadamard_finite_part_integral

Or on the complex plane

https://en.m.wikipedia.org/wiki/Kramers-Kronig_relations