Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$ Here is another interesting integral inequality :
$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$
According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference is so small.

Maybe this is easier:
$$\eqalign{
  & \int\limits_0^1 {\frac{{{x^4}\log x}}{{{x^2} - 1}}dx}  = \int\limits_0^1 {\frac{{\left( {{x^4} - 1} \right)\log x}}{{{x^2} - 1}}dx}  + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx}   \cr 
  &  = \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx}  + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx}  \cr} $$

$$ = \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx}  = \left[ {\left( {\frac{{{x^3}}}{3} + x} \right)\log x} \right]_0^1 - \int\limits_0^1 {\left( {\frac{{{x^2}}}{3} + 1} \right)dx}  =  - \frac{{10}}{9}$$

$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac 1 2 \int_0^1 \frac{\log x}{x-1}dx-\frac 1 2 \int_0^1 \frac{\log x}{x+1}dx$$
Now those two evaluate in terms of $\pi$, since we get 
$$-\int_0^1 \frac{\log x}{1-x}dx=-\int_0^1 \log x \sum_{k=0}^\infty x^kdx= $$
$$=-\sum_{k=0}^\infty\int_0^1  x^k\log x dx=\sum_{k=0}\frac{1}{(k+1)^2}=\frac{\pi^2}{6} $$
Since
$$\int_0^1  x^k\log x dx=-\frac{1}{(k+1)^2}$$
And for the other, we get the similar:
$$\int_0^1 \frac{\log x}{1+x}dx=\int_0^1 \log x \sum_{k=0}^\infty (-1)^kx^kdx= $$
$$=\sum_{k=0}^\infty\int_0^1 \log x  (-1)^kx^kdx=\sum_{k=0}^\infty(-1)^k\int_0^1 x^k\log x  dx=\sum_{k=0}\frac{(-1)^{k+1}}{(k+1)^2}=-\frac{\pi^2}{12}  $$
So we have that
$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac 1 2 \left( \frac{\pi^2}{6}+\frac{\pi^2}{12} \right)$$
$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac{\pi^2}{8}$$
and finally
$$I=\frac{\pi^2}{8}-\frac{10}{9}\approx 0.1225 < 0.125$$
