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.

Representing natural logarithm as an integral and changing the order of integration we obtain:
$$\ldots = \int_0^1 \frac{x^4}{x^2 - 1} \, dx \int_1^x \frac{dt}{t} = \int_0^1 \frac{dt}{t} \int_0^t \frac{x^4}{x^2 - 1} \, dx \\= \int_0^1 \frac{t + \frac{1}{3} t^3 - \tanh^{-1} t}{t}\, dt = \frac{10}{9} - \int_0^1 \frac{\tanh^{-1} t}{t} \, dt$$
So we now want to show that:
$$\frac{71}{72} = \frac{10}{9} - \frac{1}{8} \le \int_0^1 \frac{\tanh^{-1} t}{t} \, dt$$
Using Maclauirn series we have:
$$\frac{\tanh^{-1}}{t} = 1 + \frac{t^2}{3} + \frac{t^4}{5} + \ldots$$
Integrating first three terms yields: $\frac{259}{225} > \frac{71}{72}$.
Actually it turns out that $\frac{\tanh^{-1} t}{t} \approx 1$ is enough.
Added
Alternatively we can write:
$$\int_0^1 \frac{\tanh^{-1} t}{t} dt = \int_0^1 \frac{dt}{t} \int_0^t \frac{dx}{1-x^2} \ge \int_0^1 \frac{dt}{t} \int_0^t dx = \int_0^1 dt = 1 \ge \frac{71}{72}$$
