\(\int \frac{\log x}{(1+\log x)^{2}}dx=\)

\(\frac{x}{1-\log x}+k\) \(\frac{x}{1+\log x}+k\) \(\frac{1+\log x}{2}+k\) \(\frac{x}{1+\log x}-k\)

\(\frac{x}{1+\log x}+k\)

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