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?
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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}}} &
\,\,\,\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] & =
\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] & = \color{#f00}{\infty}
\end{align}

Even the Cauchy Principal Value $\underline{diverges}$.

A: 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
