Solution
= Solution
For $x>0$, use <integration by parts> with $u=\ln x$ and $dv=dx$. Then
$$
\boxed{\int\ln x\,dx=x\ln x-\int1\,dx=x\ln x-x+C.}
$$
Differentiating the answer checks cancellation of the extra constant term from $(x\ln x)'$.
= Solution
For $x>0$, use <integration by parts> with $u=\ln x$ and $dv=dx$. Then
$$
\boxed{\int\ln x\,dx=x\ln x-\int1\,dx=x\ln x-x+C.}
$$
Differentiating the answer checks cancellation of the extra constant term from $(x\ln x)'$.