<s>[INST] 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.
 [/INST] According to Maple, your integral is $\dfrac{\pi^2}{8} - \dfrac{10}{9}$, so your inequality becomes $\pi < \sqrt{89}/{3}$.  In fact, an antiderivative is
$$F(x) = \dfrac{x^3 \ln(x)}{3} - \dfrac{x^3}{9} + x \ln(x) - x - \dfrac{\ln(x) \ln(x+1)}{2} - \dfrac{\text{dilog}(x)+\text{dilog}(x+1)}{2}$$
More generally, for $p > -1$ $$\int_0^1 \dfrac{x^p \ln(x)}{x^2-1}\ dx = \dfrac{\Psi(1,(p+1)/2)}{4}$$
where for even integers $p=2n$, $$\Psi(1,n+1/2)  = \sum_{k=n+1}^\infty \dfrac{4}{(2k-1)^2}$$
while for odd integers $p=2n-1$,
$$\Psi(1,n) =  \sum_{k=n}^{\infty} \dfrac{1}{k^2}$$ 
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