Note that the integrand is an even function of $x$, so we will compute the integral of half the integrand over the whole real line.
Using partial fractions, we get $ \frac{1/2}{x(1-x^2)}=\frac{1/2}{x}+\frac{1/4}{1-x}-\frac{1/4}{1+x} $ Since the singularities are removable, we can use the contour $\gamma$ from $-\infty-i\epsilon$ to $+\infty-i\epsilon$ instead of $\mathbb{R}$. The key step is to break up the integral into two along two closed contours
$\gamma_+$ which goes from $-N-\frac iN$ to $+N-\frac iN$ then counterclockwise around the semicircle centered at $-\frac iN$ from $+N-\frac iN$ back to $-N-\frac iN$
$\gamma_-$ which goes from $-N-\frac iN$ to $+N-\frac iN$ then clockwise around the semicircle centered at $-\frac iN$ from $+N-\frac iN$ back to $-N-\frac iN$
$ \frac{1}{2i}\oint_{\gamma_+}\left(\frac{1/2}{z}+\frac{1/4}{1-z}-\frac{1/4}{1+z}\right)e^{i\pi z}\mathrm{d}z-\frac{1}{2i}\oint_{\gamma_-}\left(\frac{1/2}{z}+\frac{1/4}{1-z}-\frac{1/4}{1+z}\right)e^{-i\pi z}\mathrm{d}z $ As $N\to\infty$, the contribution from the semi-circular parts vanishes and we are left with the integral along $\gamma$ of $\frac{1/2}{x(1-x^2)}\sin(\pi x)=\frac{1/2}{x(1-x^2)}\dfrac{e^{i\pi x}-e^{-i\pi x}}{2i}$.
There are no singularities inside $\gamma_-$, so that integral is $0$. Thus, the whole integral boils down to $ \frac{1}{2i}\oint_{\gamma_+}\left(\frac{1/2}{z}+\frac{1/4}{1-z}-\frac{1/4}{1+z}\right)e^{i\pi z}\mathrm{d}z $ Summing up the residues at $-1,0,\text{and }1$ yields $\dfrac{1}{4i}2\pi i+\dfrac{1}{8i}2\pi i+\dfrac{1}{8i}2\pi i=\pi$.
Thus, $ \int_0^\infty\frac{\sin(\pi x)}{x(1-x^2)}\mathrm{d}x=\pi $