<s>[INST] Closed form for $\int_ 0^{1/2} {\frac {x} {3 +4 x^2}\ln\frac {\ln\left (1/2 +x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $ Find the integral $$\int_ 0^{1/2} {\frac {x} {3 + 4 x^2}\ln\frac {\ln\left (1/2 + x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $$
Wolfram | Alpha tells me that the answer is $ - 0.0951875 ... $,
and its possible closed form is $ - \frac18\ln2\ln3$. But I don't know how to do it.
Thanks in advance.
 [/INST] After integration by parts we have
\begin{gather*}
 I=\int_{0}^{1/2}\frac{x}{3+4x^2}\ln\dfrac{\ln(1/2+x)}{\ln(1/2-x)}\,\mathrm{d}x= \underbrace{\left[\dfrac{1}{8}\ln\left(\frac{3+4x^2}{4}\right)\ln\dfrac{\ln(1/2+x)}{\ln(1/2-x)}\right]_{0}^{1/2}}_{=0}-\\[2ex]
 \dfrac{1}{8}\int_{0}^{1/2}\ln\left(\frac{3+4x^2}{4}\right)\left(\dfrac{1}{\ln(1/2+x)(1/2+x)}+\dfrac{1}{\ln(1/2-x)(1/2-x)}\right)\,\mathrm{d}x= \\[2ex]
 -\dfrac{1}{8}\int_{-1/2}^{1/2}\ln\left(\frac{3+4x^2}{4}\right)\dfrac{1}{\ln(1/2+x)(1/2+x)}\,\mathrm{d}x.
 \end{gather*}
Now substitute $1/2+x=y$. We get
\begin{equation*}
  I= -\dfrac{1}{8}\int_{0}^{1}\dfrac{\ln(y^2-y+1)}{y\ln(y)}\,\mathrm{d}y.
 \end{equation*}
Finally substitute $y=e^{-t}$. Then
\begin{equation*}
  I=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{\ln(e^{-2t}-e^{-t}+1)}{t}\,\mathrm{d}t=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{\ln(e^{-3t}+1)-\ln(e^{-t}+1)}{t}\,\mathrm{d}t.
   \end{equation*}
But this is a Frullani integral. See
https://en.wikipedia.org/wiki/Frullani_integral
Put $f(t)=\ln(e^{-t}+1)$. Then
\begin{equation*}
    I=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{f(3t)-f(t)}{t}\,\mathrm{d}t=\dfrac{1}{8}(f(\infty)-f(0))\ln\left(\dfrac{3}{1}\right) =-\dfrac{1}{8}\ln(2)\ln(3).
   \end{equation*}
</s>