TITLE: How to show $\int_0^1 \frac{\ln\frac{1+a}{1+x}}{a-\frac1a{x^2}}dx=\frac{\pi^2}{24}-\frac14\text{Li}_2\left[\left(\frac{a-1}{a+1}\right)^2\right] $
QUESTION [5 upvotes]: I came across this polylogrithmic integral that evaluates to a close form of dilogarithmic value
$$\int_0^1 \frac{\ln\frac{1+a}{1+x}}{a-\frac1a{x^2}}dx=\frac{\pi^2}{24}-\frac14\operatorname{Li}_2\bigg[\bigg(\frac{a-1}{a+1}\bigg)^2\bigg]
$$
for $a>1$. So far, I am not able to derive it. Knowing the form of the result, I expect to utilize the identity
$$\operatorname{Li}_2(z)+\operatorname{Li}_2(-z)= \frac12\operatorname{Li}_2(z^2)$$
But, I was not able to manipulate the integrand to fit it into the desired sum of dilogarithm integrals. It may not be as straightforward as I expected.

REPLY [6 votes]: $$I=\int_0^1 \frac{\ln\left(\frac{1+a}{1+ax}\right)}{a-\frac{1}{a}x^2}dx\overset{\large x\to\, a\frac{1-x}{1+x}}=\frac12\int_k^1\frac{\ln\left(\frac{1+x}{1-kx}\right)}{x}dx=I(k); \ k=\frac{a-1}{a+1}$$
$$\frac{\partial}{\partial k}I(k)=\frac{\ln(1-k)}{2k}+\frac12\int_k^1 \frac{1}{1-kx}dx=\frac{\ln(1-k^2)}{2k}$$
$$\Rightarrow I=\frac12\int_{-1}^k\frac{\ln(1-x^2)}{x}dx=-\frac14\operatorname{Li}_2(x^2)\bigg|_{-1}^k=\frac{\pi^2}{24}-\frac14\operatorname{Li}_2(k^2)$$