Contour integral of $\int_{-\infty}^{\infty}\frac{xdx}{(x-1)^{4}-1}$ I'm having a little trouble solving this contour integral. I've found the singularities to be at 0, 2, and $1\pm i$. I'm not quite sure how to evaluate the integral when there are two poles on the contour instead of just one. Any help is appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Assuming the question asks for the Principal Value $\pars{~\mrm{P.V.}~}$:

\begin{align}
&\mrm{P.V.}\int_{-\infty}^{\infty}{x\,\dd x \over \pars{x - 1}^{4} - 1}
\\[5mm] \stackrel{\mrm{def.}}{=} &\
\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{-\infty}^{-\epsilon}{x\,\dd x \over \pars{x - 1}^{4} - 1} +
\int_{\epsilon}^{2 - \epsilon}{x\,\dd x \over \pars{x - 1}^{4} - 1} +
\int_{2 + \epsilon}^{\infty}{x\,\dd x \over \pars{x - 1}^{4} - 1}}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{-\infty}^{-1 - \epsilon}{x + 1 \over x^{4} - 1}\,\dd x +
\int_{-1 + \epsilon}^{1 - \epsilon}{x + 1 \over x^{4} - 1}\,\dd x +
\int_{1 + \epsilon}^{\infty}{x + 1 \over x^{4} - 1}\,\dd x}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{-\infty}^{-1 - \epsilon}{\dd x \over \pars{x - 1}\pars{x^{2} + 1}}\,\dd x +
2\int_{0}^{1 - \epsilon}{\dd x \over x^{4} - 1} -
\int_{0}^{1/\pars{1 + \epsilon}}{x \over \pars{x - 1}\pars{x^{2} + 1}}\,\dd x}
\\[1cm] = &\
-\int_{1}^{\infty}{\dd x \over \pars{x + 1}\pars{x^{2} + 1}}
\\[5mm] + &\
\lim_{\epsilon \to 0^{+}}\braces{%
\int_{0}^{1 - \epsilon}\bracks{%
{2 \over x^{4} - 1} - {x \over \pars{x - 1}\pars{x^{2} + 1}}}\,\dd x -
\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}}
{x \over \pars{x - 1}\pars{x^{2} + 1}}\,\dd x}
\\[1cm] & =
-\int_{1}^{\infty}{\dd x \over \pars{x + 1}\pars{x^{2} + 1}} -
\int_{0}^{1}{x + 2 \over \pars{x + 1}\pars{x^{2} + 1}}\,\dd x
\\[3mm] - & \ \underbrace{%
\lim_{\epsilon \to 0^{+}}\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}}
{x \over \pars{x - 1}\pars{x^{2} + 1}}\,\dd x}_{\ds{=\ 0}} =
\bbx{-\ds{\pi \over 2}}
\end{align}

I left to you the integral evaluations and the limit evaluation.

A: $$\text{PV}\int_{-\infty}^{+\infty}\frac{x}{(x-1)^4-1}\,dx = \text{PV}\int_{-\infty}^{+\infty}\frac{x+1}{x^4-1}\,dx = \text{PV}\int_{-\infty}^{+\infty}\frac{dx}{(x-1)(x^2+1)} $$
equals, by partial fraction decomposition,
$$ \frac{1}{2}\,\text{PV}\int_{-\infty}^{+\infty}\frac{dx}{x-1}-\frac{1}{2}\,\text{PV}\int_{-\infty}^{+\infty}\frac{x+1}{x^2+1}\,dx = \color{red}{-\frac{\pi}{2}}.$$
