In short, you were right. In long, I'm going to do it here myself.
This is the contour I will be using
If we integrate around our contour and apply the residue theorem, we get the following
$ \begin{array}{l}
\oint _{C}\frac{1}{z\left( z^{2} +1\right)} dz=\\
2\pi i\cdot Res\left(\frac{1}{z\left( z^{2} +1\right)} ,\ z=0\right) +2\pi i\cdot Res\left(\frac{1}{z\left( z^{2} +1\right)} ,\ z=i\right) =\pi i,\\
\\
\oint _{C}\frac{1}{z\left( z^{2} +1\right)} dz=\int\limits ^{\infty }_{-\infty }\frac{1}{x\left( x^{2} +1\right)} dx+\int _{A} +\int _{B} \Longrightarrow \\
\\
\int\limits ^{\infty }_{-\infty }\frac{1}{x\left( x^{2} +1\right)} dx=\pi i-\int _{A} -\int _{B}\\
\\
\int _{A} =0,\ through\ triangle\ inequalities\ ( ask\ me\ if\ anyone\ wants\ futher\ details)\\
\\
\int _{B} =\pi i\cdot Res\left(\frac{1}{z\left( z^{2} +1\right)} ,\ 0\right) =\pi i\\
\\
\therefore \int\limits ^{\infty }_{-\infty }\frac{1}{x\left( x^{2} +1\right)} dx=\pi i-\pi i=0\\
\blacksquare
\end{array}$
I believe the black square represents the phrase QED.
For an approach not using contour integration/complex analysis.
When a function is odd, it has this property $f(x)=-f(-x)$. The integrand of your integral is odd. The property of odd integrands can be taken advantage of like so
$ \begin{array}{l}
\int\limits ^{\infty }_{-\infty }\frac{1}{x\left( x^{2} +1\right)} dx=\int\limits ^{\infty }_{0}\frac{1}{x\left( x^{2} +1\right)} dx+\underbrace{\int\limits ^{0}_{-\infty }\frac{1}{x\left( x^{2} +1\right)} dx}_{Make\ the\ subsitution\ here}\\
make\ a\ substitution,\ \ x\mapsto -x\Longrightarrow \\
\int\limits ^{\infty }_{-\infty }\frac{1}{x\left( x^{2} +1\right)} dx=\int\limits ^{\infty }_{0}\frac{1}{x\left( x^{2} +1\right)} dx-\int\limits ^{\infty }_{0}\frac{1}{x\left( x^{2} +1\right)} dx=0
\end{array}$
In general, you will find this
For an odd function
$f( x)
\\
\int\limits ^{k}_{-k} f( x) dx=0$
Also as a side note
For even functon $f(x)$
$f(x)=f(-x)$ and also
$\int\limits^{k}_{-k} f(x) dx=2\cdot\int\limits^{k}_{0} f(x) dx$
If anything needs a further explanation then please ask. Thanks!
Best Answer
Write $\sin{x} = (e^{i x}-e^{-i x})/(2 i)$. Then consider the integral
$$PV \oint_{C_{\pm}} dx \frac{e^{\pm i z}}{z (z^2-2 z+2)} $$
where $C_{\pm}$ is a semicircular contour of radius $R$ in the upper/lower half plane with a semicircular detour into the upper/lower half plane of radius $\epsilon$. For $C_{+}$, we have
$$PV \oint_{C_{+}} dz \frac{e^{i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i R e^{i \theta}}}{R e^{i \theta} (R^2 e^{i 2 \theta} - 2 R e^{i \theta}+2)} $$
For $C_-$, we have
$$PV \oint_{C_{-}} dz \frac{e^{-i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{-i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{-\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{-i x}}{x (x^2-2 x+2)}- i R \int_0^{\pi} d\theta \, e^{-i \theta} \frac{e^{-i R e^{-i \theta}}}{R e^{-i \theta} (R^2 e^{-i 2 \theta} - 2 R e^{-i \theta}+2)} $$
In both cases, we take the limits as $R \to \infty$ and $\epsilon \to 0$. Note that, in both cases, the respective fourth integrals have a magnitude bounded by
$$\frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi}\le \frac{\pi}{R^3}$$
The respective second integrals of $C_{\pm}$, on the other hand, become equal to $\mp i \frac{\pi}{2} $. Thus,
$$PV \oint_{C_{\pm}} dz \frac{e^{\pm i z}}{z (z^2-2 z+2)} = PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2}$$
On the other hand, the respective contour integrals are each equal to $\pm i 2 \pi$ times the sum of the residues of the poles inside their contours. (For $C_-$, there is a negative sign because the contour was traversed in a clockwise direction.) The poles of the denominator are at $z_{\pm}=1 \pm i$. Thus,
$$PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2} = \pm i 2 \pi \frac{e^{\pm i (1 \pm i)}}{(1 \pm i) (2) (\pm i)} $$
Taking the difference between the two results and dividing by $2 i$, we get that
$$\int_{-\infty}^{\infty} dx \frac{\sin{x}}{x (x^2-2 x+2)} = \frac{\pi}{2} \left (1+\frac{\sin{1}-\cos{1}}{e} \right ) $$
Note that we may drop the $PV$ because the difference between the integrals removes the pole at the origin.