calculation and analysis – Simple definite integral returns not evaluated

The integral below returns unrated

To integrate[ x/(1+z^x), {x,0,Infinity}, Assumptions->z > 1]  

But a simple redefinition of the independent variables gives

To integrate[- x/(1+1/z^x), {x,-Infinity,0}, 
Assumptions-> z > 1] = pi ^ 2 / (12 Log[z]^ 2)