Integrating by Parts to Find \(\int \ln x\, dx\)

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 \[ \int \ln x\,dx= {{x}}\cdot {{\ln x}}-\int {{x}}\cdot {{\frac{1}{x}\,dx}}\,=x\,\ln x-\int dx=x\,\ln x-x+C \]