∫ (log x)/((1+log x)^2)dx=

$x=e^y⇒dx=e^ydy$

$I=\int\frac{\log xdx}{(1+\log x)^2}=\int\frac{ye^ydy}{(1+y)^2}=\int e^y\frac{y+1-1}{(y+1)^2}dy$

$=\int e^y\left(\frac{1}{y+1}-\frac{1}{(y+1)^2}\right)dy=\frac{e^y}{y+1}+C$

$=\frac{x}{\log x+1}+C$