Solution
= Solution
For real powers work on $x>0$. Apply <integration by parts> with $u=\ln x$ and $dv=x^m dx$, so $v=x^{m+1}/(m+1)$ when $m\ne-1$. Then
$$
\boxed{\int x^m\ln x\,dx
=\frac{x^{m+1}\ln x}{m+1}-\frac{x^{m+1}}{(m+1)^2}+C.}
$$
= Solution
For real powers work on $x>0$. Apply <integration by parts> with $u=\ln x$ and $dv=x^m dx$, so $v=x^{m+1}/(m+1)$ when $m\ne-1$. Then
$$
\boxed{\int x^m\ln x\,dx
=\frac{x^{m+1}\ln x}{m+1}-\frac{x^{m+1}}{(m+1)^2}+C.}
$$