An integral from $0$ to $\infty$ with $\mathrm dv/v$ in it tends to be susceptible to a substitution of the form $u=\lambda v$, which leaves all of that invariant. In the present case, you can write $u=xv$ to get
$ \int_0^\infty \left( \frac{1}{ 1 + x^\alpha \vert v-1 \vert^\alpha} - \frac{1}{ 1 + x^\alpha \vert v+1 \vert^\alpha} \right) \frac{\mathrm d v}{v} = \int_0^\infty \left( \frac{1}{ 1 + \vert u-x \vert^\alpha} - \frac{1}{ 1 + \vert u+x \vert^\alpha} \right) \frac{\mathrm d u}{u}\;. $
The integrand is even, so you can simplify things by calculating the integral from $-\infty$ to $\infty$ instead.
The denominators are of the form $z^\alpha+1$, which for odd $\alpha$ factorizes into $\prod_i(z+z_i)$, where $z_i$ are the $\alpha$-th roots of unity. So we have
$\frac12\int_{-\infty}^\infty \left( \frac{1}{\prod_i(\vert u-x \vert+z_i)} - \frac{1}{\prod_i(\vert u+x \vert+z_i)} \right) \frac{\mathrm d u}{u}\;.$
The rest is an exercise in partial fractions, with the two complex conjugate solutions combining into a quadratic denominator; I think you'll need to resolve the absolute values first; let me know if you want me to write it out further.
P.S.: You can further simplify things by substituting $t=u-x$ in the first term and $t=u+x$ in the second; that yields
$\frac12\int_{-\infty}^\infty \frac{1}{\prod_i(\vert t \vert+z_i)}\left(\frac1{t+x}-\frac1{t-x} \right) \mathrm d t\;. $