Aus Online Mathematik Brückenkurs 2

It is simpler to investigate the integral if we write it as

\displaystyle \int{\ln x\centerdot \frac{1}{x}\,dx},

The derivative of \displaystyle \ln x is \displaystyle \frac{1}{x}, so if we choose \displaystyle u=\ln x, the integral can be expressed as

\displaystyle \int{u\centerdot {u}'\,dx}

Thus, it seems that \displaystyle u=\ln x is a useful substitution,

\displaystyle \begin{align} & \int{\ln x\centerdot \frac{1}{x}\,dx}=\left\{ \begin{matrix} u=\ln x \\ du=\left( \ln x \right)^{\prime }\,dx=\frac{1}{x}\,dx \\ \end{matrix} \right\} \\ & =\int{u\,du=\frac{1}{2}u^{2}+C} \\ & =\frac{1}{2}\left( \ln x \right)^{2}+C \\ \end{align}