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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.

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    Calculate the residues ... Cauchy's Residue theorem etc... – 2017-02-07
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    Maybe, you're asking for the $\textit{Principal Value}$. – 2017-02-07
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    @JohnPage Just to clarify there is an $x$ in the numerator ? ... and the in denominator, that will cancel so your integral can be rewritten as $\int _{-\infty}^{\infty} \frac{dx}{(x-2)(x^2-2x+2)}$ So there is only one singularity on the contour. – 2017-02-07
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    There is a subtle but important difference between $\text{PV}\int_{-\infty}^{+\infty}\frac{dx}{x}=0$ and $\int_{-\infty}^{+\infty}\frac{dx}{x}$, that simply does not exist. The same applies to the question. – 2017-02-07

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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.

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    @JackD'Aurizio I checked it with $\texttt{Mathematica}$. $$ \texttt{Integrate[x/((x - 1)^4 - 1), {x, -Infinity, Infinity}, PrincipalValue -> True]} $$ – 2017-02-07
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    Sorry, my bad: I previously simplified $\frac{x+1}{x^2-1}$ as $\frac{1}{x+1}$ instead of $\frac{1}{x-1}$, now fixed. – 2017-02-07
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    @JackD'Aurizio It still works: $-\pi/2$. Thanks. – 2017-02-07
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    Sure, sure, you are right. It was me not double-checking my computations. (+1) to you. – 2017-02-07
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$$\text{PV}\int_{-\infty}^{+\infty}\frac{x}{(x-1)^4-1}\,dx = \text{PV}\int_{-\infty}^{+\infty}\frac{x+1}{x^4-1}\,dx = \text{PV}\int_{-\infty}^{+\infty}\frac{dx}{(x-1)(x^2+1)} $$ equals, by partial fraction decomposition, $$ \frac{1}{2}\,\text{PV}\int_{-\infty}^{+\infty}\frac{dx}{x-1}-\frac{1}{2}\,\text{PV}\int_{-\infty}^{+\infty}\frac{x+1}{x^2+1}\,dx = \color{red}{-\frac{\pi}{2}}.$$

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    Ah, good use shifting the integral and PFD. – 2017-02-07