Evaluate the integral :
let $\displaystyle \alpha > 0$ and put $\displaystyle x = \frac{1}{t}.$ then we get $\displaystyle \int \frac {dx}{x(1 + x^{\alpha})} = - \int \frac {t^{\alpha - 1}}{1+t^{\alpha}} \ dt = - \frac{1}{\alpha} \ln |1 + t^{\alpha}| + c=-\frac{1}{\alpha} \ln |1 + x^{-\alpha}| + c.$