Let
$\displaystyle \displaystyle I(a,m)=\int_0^{\infty}\frac{x^{a-1}}{1+x^m},\quad m>1,\;\; 0<a<m$
Then, let $\displaystyle x^m=u$ to get that
$\displaystyle \displaystyle I(a,m)=\int_0^{\infty}\frac{u^{\frac{a-1}{m}}}{1+u}\frac{du}{m u^{1-\frac{1}{m}}}=\frac{1}{m}\int_0^{\infty}\frac{u^{\ frac{a}{m}-1}}{1+u}\text{ }du$
Note though that since $\displaystyle \displaystyle 0<\frac{a}{m}<1$ we may apply the methods in the previous problem to conclude that
$\displaystyle \displaystyle I(a,m)=\frac{1}{m}\frac{\pi}{\sin\left(\frac{\pi a}{m}}\right)}$
Thus,
$\displaystyle \displaystyle \int_0^{\infty}\frac{x^{a-1}\log(x)}{1+x^m}\text{ }dx=\frac{\partial}{\partial a}I(a,m)=-\left(\frac{\pi}{m}\right)^2\cot\left(\frac{\pi a}{m}\right)\csc\left(\frac{\pi a}{m}\right)$