TITLE: Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$
QUESTION [10 upvotes]: 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.

REPLY [11 votes]: You can actually just evaluate the integral explicitly. You can divide $x^2 -1$ into $x^4$ and get
$$\frac{x^4}{x^2 - 1} = x^2 + 1 + \frac {1}{x^2 - 1}$$
So the integral is the same as
$$\int_0^1 (x^2 + 1)\log(x)\,dx + \int_0^1 \frac{\log(x)}{x^2 - 1}\,dx $$
The second integral is related to the famous dilogarithm integral, and as explained in Peter Tamaroff's answer can be evaluated to $\frac{\pi^2}{8}$. For the first term, just integrate by parts; you get
$$({x^3 \over 3} + x)\log(x)\big|_{x = 0}^{x =1} - \int_0^1 ({x^2 \over 3} + 1)\,dx$$
The first term vanishes, while the second term is $-{10 \over 9}$. So the answer is just ${\pi^2 \over 8} - {10 \over 9}$ which is less than ${1 \over 8}$.

A way of doing the whole integral in one fell swoop occurs to me. Note that ${\displaystyle {1 \over 1 - x^2} = \sum_{n=0}^{\infty} x^{2n}}$. So the integral is 
$$-\sum_{n = 0}^{\infty} \int_0^1 x^{2n + 4}\log(x)\,dx$$
$$= -\sum_{m = 2}^{\infty} \int_0^1 x^{2m}\log(x)\,dx$$
Integrating this by parts this becomes
$$\sum_{m = 2}^{\infty} \int_0^1 {x^{2m} \over 2m + 1}$$
$$= \sum_{m = 2}^{\infty} {1 \over (2m + 1)^2}$$
This is the sum of the reciprocals of the odd squares starting with $5$. The sum of the reciprocals of all odd squares is ${\pi^2 \over 8}$, so one subtracts off $1 + {1 \over 9} = {10 \over 9}$. Hence the result is $ {\pi^2 \over 8} - {10 \over 9} $.