Derive the formula $$\bbox[5px, border:1px solid black, #F9F7ED]{ \int \ln x\,dx=x\ln x-x+C }$$

Solution We use the integration by parts formula with $$\begin{equation*} \hbox{ }u=\ln x\qquad \hbox{and}\qquad dv=dx \end{equation*}$$

Then $$du=\frac{1}{x}\,dx\qquad {\rm and}\qquad v=\int dx=x \ $$

Now $$\int \ln x\,dx= {{x}}\cdot {{\ln x}}-\int {{x}}\cdot {{\frac{1}{x}\,dx}}\,=x\,\ln x-\int dx=x\,\ln x-x+C$$