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

$\begin{array}{l}\text{ Let }\log x=t\Rightarrow x=e^t\\ \int e^t\left[\frac{1}{1+t}+\frac{-1}{{\left(1+t^{ }\right)}^2}\right]dt\\ f\left(t\right)=\frac{1}{1+t}\\ =e^t\frac{1}{1+t}+c=\frac{x}{1+\log x}+c\end{array}$