Evaluate $\lim\limits_{x\to\infty}x\left(1+\frac2\pi\int_{0}^{1}\tan^{-1}\left(t^x-t^{-x}\right)\,dt\right)$ To find $\lim_{x \to \infty} x\left( 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt \right)$ I tried the L'Hopital's rule but it cannot solve the problem.
Here is my approach:
$$\lim_{x \to \infty} \frac{ 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt }{\frac{1}{x}}=\lim_{x \to \infty}\frac{\int_{0}^{1}\frac{2\ln{t}\;{\cosh{(x\ln{t})}}}{1+4\sinh^2{(x\ln{t})}}dt}{{\frac{-1}{x^2}}}=\frac{0}{0}$$
Still $\frac{0}{0}$ and second L'Hopital's won't help. Any help is appreciated.
 A: Partial evaluation of @Svyatoslav's log integral:
$$\begin{align*}
J &= \int_0^1 \frac{1+t^2}{1-t^2+t^4} \ln(t) \, dt \\[1ex]
&= -\frac{e^{i\pi/3}}{1-e^{i\pi/3}} \int_0^1 \frac{\ln(t)}{t^2-e^{i\pi/3}} \, dt - \frac{e^{-i\pi/3}}{1-e^{-i\pi/3}} \int_0^1 \frac{\ln(t)}{t^2-e^{-i\pi/3}} \, dt \tag{1} \\[1ex]
&= e^{i\pi/3} \int_0^1 \frac{\ln(t)}{1-e^{-i\pi/3}t^2} \, dt + e^{-i\pi/3} \int_0^1 \frac{\ln(t)}{1-e^{i\pi/3}t^2} \, dt \\[1ex]
&= \sum_{n=0}^\infty \left( e^{-i(n-1)\pi/3} + e^{i(n-1)\pi/3}\right) \int_0^1 t^{2n} \ln(t) \, dt \tag{2} \\[1ex]
&= -2 \sum_{n=0}^\infty \frac{\cos\left(\frac{(n-1)\pi}3\right)}{(2n+1)^2} \tag{3} \\[1ex]
&= - \sum_{n=0}^\infty \frac{\cos\left(\frac{n\pi}3\right)+ \sqrt3 \sin\left(\frac{n\pi}3\right)}{(2n+1)^2}
\end{align*}$$
The remaining sum can be expressed in terms of hypergeometric functions, such that
$$J = e^{i2\pi/3} {}_{3}F_{2} \left(\left.\begin{array}{c|c}\frac12,\frac12,1\\\frac32,\frac32\end{array}\right\vert e^{i\pi/3}\right) + e^{-i2\pi/3} {}_{3}F_{2} \left(\left.\begin{array}{c|c}\frac12,\frac12,1\\\frac32,\frac32\end{array}\right\vert e^{-i\pi/3}\right)$$
but it's not immediately clear to me how to simplify this...


*

*$(1)$ : partial fractions

*$(2)$ : $\displaystyle \frac1{1-x}=\sum_{n\ge0}x^n$

*$(3)$ : integrate by parts and simplify


Update: A more complete evaluation of the integral
Taking the sum we get after integrating in line $(2)$, preserving the exponential form, we get
$$-\sum_{n=0}^\infty \frac{(\bar z)^{n-1} + z^{n-1}}{(2n+1)^2}$$
where $z={e^{i\pi/3}}$.
We have for $|x|<1$, using the fundamental theorem of calculus,
$$\begin{align*}
f(x) &= \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2} \\[2ex]
\implies f'(x) &= \sum_{n=0}^\infty \frac{x^{2n}}{2n+1} = \frac{\operatorname{artanh}(x)}{x} \\[2ex]
\implies f(x) &= \int_0^x \frac{\operatorname{artanh}(y)}{y} \, dy \\[1ex]
&= \frac12 \int_0^x \frac{\ln(1+y) - \ln(1-y)}{y} \, dy \\[1ex]
&= \operatorname{Li}_2(x) - \frac14 \operatorname{Li}_2(x^2) \\[2ex]
\implies \sum_{n=0}^\infty \frac{x^{n-1}}{(2n+1)^2} &= \frac{f(\sqrt x)}{x\sqrt x}
\end{align*}$$
and hence another form for the integral in terms of the dilogarithm,
$$J = - \frac{f(\sqrt z)}{z\sqrt z} - \frac{f(\sqrt{\bar z})}{\bar z \sqrt{\bar z}}$$
where I assume a branch cut along the negative real axis for the square root. This reduces further but I'm not as well-versed in the dilogarithm identities as I could be. Mathematica uses the same branch, and evaluates either sum with $z$ or $\bar z$ as $\dfrac23G \pm i \dfrac{\pi^2}{12}$, hence $J=-\dfrac43G$.
In[]:= f[x_] := PolyLog[2, x] - 1/4 PolyLog[2, x^2];
In[]:= f[E^(I \[Pi]/6)]/(E^(I \[Pi]/3) E^(I \[Pi]/6)) // FunctionExpand // Expand

Out[]= (2 Catalan)/3 - (I \[Pi]^2)/12

