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.
 A: 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:
$$\int_0^1 \frac{\tanh^{-1} t}{t} dt = \int_0^1 \frac{dt}{t} \int_0^t \frac{dx}{1-x^2} \ge \int_0^1 \frac{dt}{t} \int_0^t dx = \int_0^1 dt = 1 \ge \frac{71}{72}$$
A: 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
$$I=\frac{\pi^2}{8}-\frac{10}{9}\approx 0.1225 < 0.125$$
A: 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 $p=2n-1$,
$$\Psi(1,n) =  \sum_{k=n}^{\infty} \dfrac{1}{k^2}$$ 
A: 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} $.
