Aus Online Mathematik Brückenkurs 2

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

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

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

\displaystyle \int u\cdot u'\,dx\,\textrm{.}

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

\displaystyle \begin{align} \int \ln x\cdot\frac{1}{x}\,dx &= \left\{\begin{align} u &= \ln x\\[5pt] du &= (\ln x)'\,dx = (1/x)\,dx \end{align}\right\}\\[5pt] &= \int u\,du\\[5pt] &= \frac{1}{2}u^{2} + C\\[5pt] &= \frac{1}{2}(\ln x)^2 + C\,\textrm{.} \end{align}