Solution
= Solution
<Integration by parts> with $u=\ln x$ and $dv=x\,dx$ gives
$$
\boxed{\int x\ln x\,dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C,\qquad x>0.}
$$
The remaining <integral> is $\frac12\int x\,dx$.
= Solution
<Integration by parts> with $u=\ln x$ and $dv=x\,dx$ gives
$$
\boxed{\int x\ln x\,dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C,\qquad x>0.}
$$
The remaining <integral> is $\frac12\int x\,dx$.