On the proof of inequality $\int_0^1 x^\alpha \ln(x) dx \leq \frac{1}{\alpha+1} \ln\left(\frac{\alpha+1}{\alpha+2}\right)$ Other than evaluating the definite integral and comparing it to the right side of the inequality, is there another way to show this? Thanks: 
$\displaystyle \int_0^1 x^\alpha \ln x  ~dx \leq \displaystyle \frac{1}{\alpha+1}ln \frac{\alpha+1}{\alpha+2}$ for $\alpha>0$.
 A: Considering the lhs, using one integration by parts $$\int x^a \ln x  ~dx=\frac{x^{a+1} ((a+1) \log (x)-1)}{(a+1)^2}$$ So $$\int_0^1 x^a \ln x  ~dx=-\frac{1}{(a+1)^2}$$ provided that $\Re(a)>-1$.
So, we have to show that $$-\frac{1}{(a+1)^2}\leq \frac{1}{a+1}\log\left( \frac{a+1}{a+2}\right)$$ that is to say $$-1 \leq (a+1)\,\log\left( \frac{a+1}{a+2}\right)$$ We can rewrite is as $$-\frac 1{a+1}\leq -\log\left(1+ \frac 1{a+1}\right)$$
Can you take it from here ?
A: This answer is essentially the same as Claude Leibovici's. Note
\begin{eqnarray}
&&\int_0^1 x^\alpha \ln x  ~dx -\frac{1}{\alpha+1}\ln \frac{\alpha+1}{\alpha+2} \\
&=&\int_0^1 x^\alpha \ln x  ~dx -\ln \frac{\alpha+1}{\alpha+2}\int_0^1x^\alpha ~dx\\
&=&\int_0^1  x^\alpha \left(\ln x +\ln \frac{\alpha+2}{\alpha+1}\right)dx\\
&=&\int_0^1  x^\alpha \left( \ln x+ \ln\left(1+\frac{1}{\alpha+1}\right)\right)dx\\
&\le& \int_0^1  x^\alpha \left( \ln x+\frac{1}{\alpha+1}\right)dx\\
&=&0,
\end{eqnarray}
and hence
$$ \int_0^1 x^\alpha \ln x  ~dx \le\frac{1}{\alpha+1}\ln \frac{\alpha+1}{\alpha+2}.$$
Here
$$ \ln\left(1+\frac{1}{\alpha+1}\right)\le \frac1{\alpha+1}. $$
