Evaluate: $\frac{u^{2n+1}\ln(u)}{2n+1}\big|^{u=1}_{u=0}$ at $u=0$ Problem: Evaluate:

$$\dfrac{u^{2n+1}\ln(u)}{2n+1}\bigg|^{u=1}_{u=0}$$

This actually came up when I tried to solve the Integral $$\int_0^1 u^{2n}\ln(u) du$$ by using IBP. The problem is that I am unable to evaluate this at $u=0$.$$$$ Would somebody please be so kind as to help me evaluate $\dfrac{u^{2n+1}\ln(u)}{2n+1}\bigg|^{u=1}_{u=0}?$ I would be indeed very grateful for your assistance. Many thanks in advance!
 A: The evaluation at $u=1$ is not a real issue, and since:
$$ \lim_{u\to 0^+} u\log u = 0 $$
is easy to prove through a change of variable, for instance, neither the evaluation in the other endpoint is difficult.
In any case,
$$ \int_{0}^{1} u^{2\alpha}\log(u)\,du = \frac{1}{2}\frac{d}{d\alpha}\int_{0}^{1}u^{2\alpha}\,du = \frac{1}{2}\frac{d}{d\alpha}\frac{1}{2\alpha+1}=-\frac{1}{(2\alpha+1)^2}.$$
A: Using $\partial_{x} \, u^{ax} = \partial_{x} \, e^{ax \ln(u)} = a \ln(u) \, u^{ax}$ then
\begin{align}
I &= \int_{0}^{1} u^{a x} \, \ln(u) \, du = \frac{1}{a} \partial_{x} \, \int_{0}^{1} u^{ax} \, du \\
&= \frac{1}{a} \partial_{x} \left[ \frac{u^{ax+1}}{ax+1} \right]_{0}^{1} \\
&= \frac{1}{a} \partial_{x} \left( \frac{1}{ax+1} \right) \\
&= \frac{(-1)^{1}}{(ax+1)^{2}}
\end{align}
Now let $a=2$ to obtain
\begin{align}
\int_{0}^{1} u^{2n} \, \ln(u) \, du = - \frac{1}{(2n+1)^{2}}.
\end{align}
The process may be continued to obtain
\begin{align}
\int_{0}^{1} u^{2n} \, \ln^{m}(u) \, du = \frac{(-1)^{m} \, m!}{(2n+1)^{m+1}}.
\end{align}
