We can rewrite the integral as
$$\lim_{a\to\infty} \frac{1}{a}\int_0^\infty \frac{x^2+1}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right)\:dx + \int_0^\infty \frac{x}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right)\:dx$$
which we are allowed to split up since both pieces are absolutely convergent. Then notice that
$$ \frac{x}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right) = \frac{1}{x^2}\cdot \frac{\frac{1}{x}}{1+\frac{1}{x^4}}\tan^{-1}\left(\frac{1}{x}\right) = \frac{d}{dx}\left[-\frac{1}{4}\arctan^2\left(\frac{1}{x}\right)\right]$$
therefore the integral evaluates to $\frac{\pi^2}{16}$
The denominator is of the fifth degree (after linearizing the tangent), so if there is a finite answer you will need five successive applications of L'Hospital.
The numerator is easy:
$$\sinh x-x\cosh x+\frac{x^3}3,$$
$$-x\sinh x+x^2,$$
$$-\sinh x-x\cosh x+2x,$$
$$-2\cosh x-x\sinh x+2$$
$$-3\sinh x-x\cosh x,$$
$$-4\cosh x-x\sinh x.$$
Every time, you need to check that the expression tends to zero (otherwise the limit will not exist because of the zero denominator).
For the denominator, it is really worth to rewrite
$$x^2\tan^3x=x^5\frac{\tan^3x}{x^3}$$ and take the fraction away. Then the fifth derivative is $5!$ and the requested ratio
$$-\frac1{30}.$$
For info, keeping the denominator as is, we get
$$2520x^2(\tan(x))^8+6600x^2(\tan(x))^6+3600x(\tan(x))^7+36x^2(\sec(x))^2(\tan(x))^2+\\
5772x^2(\tan(x))^4+8160x(\tan(x))^5+1200(\tan(x))^6+120x\tan(x)(\sec(x))^2+
\\120x^2(\sec(x))^2+1692x^2(\tan(x))^2+5640x(\tan(x))^3+2280(\tan(x))^4+
\\1080x\tan(x)+120(\sec(x))^2+1080(\tan(x))^2$$
and the only nonzero term is $120\sec^2x$.
Best Answer
Let's denote $\,I(x)=\displaystyle \int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt$
Integrating by part $$I(x)=t\tan^{-1}\left( t^x-t^{-x} \right)\Big|_0^1-x\int_{0}^{1}\frac{\left( t^x+t^{-x} \right)}{1+\left( t^x-t^{-x} \right)^2}dt$$ Making the substitution $t=e^{-s}$ $$=-x\int_0^\infty\frac{\left( e^{sx}+e^{-sx}\right)}{1+\left( e^{sx}-e^{-sx}\right)^2}e^{-s}ds\overset{t=sx}{=}-2\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}e^{-\frac{t}{x}}dt$$ With the accuracy up to exponentially small corrections at $x\gg1$, we can decompose the exponent: $\displaystyle e^{-\frac{t}{x}}=1-\frac{t}{x}+\frac{t^2}{2x^2}+O\Big(\frac{1}{x^3}\Big)$ $$I(x)=-2\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}dt+\frac{2}{x}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}t\,dt+O\Big(\frac{1}{x^2}\Big)$$ $$=-\frac{\pi}{2}+\frac{2}{x}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t\,dt-\frac{1}{x^2}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t^2\,dt+O\Big(\frac{1}{x^3}\Big)$$ and, therefore, the desired limit $$\lim_{x \to \infty} x\left( 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt \right)=\frac{4}{\pi}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t\,dt=0.77749...$$
(it can also be presented in the form $\displaystyle -\,\frac{2}{\pi}\int_0^1\frac{1+t^2}{1-t^2+t^4}\ln t\,dt\,$).
The numeric evaluation with WolframAlpha confirms the answer.
$\bf{Update}$
As @170231 mentioned, the last integral has a closed form: $$ J=\int_0^1\frac{1+t^2}{1-t^2+t^4}\ln t\,dt=-\,\frac{4}{3}G$$ and the desired limit is $$\boxed{\,\,\lim_{x \to \infty} x\left( 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt \right)=\frac{8G}{3\pi}=0.77749...\,\,}$$ Indeed, $$J=\int_0^1\frac{(1+t^2)^2}{1+t^6}\ln t\,dt=$$ $$\int_0^1(1-t^6+t^{12}-...)\ln t\,dt+2\int_0^1(t^2-t^8+t^{14}-...)\ln t\,dt$$ $$+\int_0^1(t^4-t^{10}+t^{14}-...)\ln t\,dt$$ $$=-1+\frac{1}{7^2}-\frac{1}{13^2}+..+2\Big(-\frac{1}{3^2}+\frac{1}{9^2}-\frac{1}{15^2}-..\Big)-\frac{1}{5^2}+\frac{1}{11^2}-\frac{1}{17^2}+...$$ Denoting the Catalan' constant $\displaystyle G=1-\frac{1}{3^2}+\frac{1}{5^2}-...$ $$J=-1+\frac{1}{3^3}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{9^2}+ ... -\frac{2}{9}G-\frac{1}{3^2}+\frac{1}{9^2}-\frac{1}{15^2}+...$$ $$=-G-\frac{2}{9}G-\frac{1}{9}G=-\,\frac{4}{3}G$$ It is interesting to note that the second term of the asymptotics $I(x)$ can also be found in a closed form: $$I(x)=I_0+\frac{1}{x}I_1+\frac{1}{x^2}I_2\,+...$$ where $$I_2=-\int_0^\infty\frac{t^2\cosh t}{1+4\sinh^2t}\,dt=-\frac{1}{2}\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}\int_{-\infty}^\infty\frac{e^{-\beta t}\cosh t}{1+4\sinh^2t}dt=-\frac{1}{2}\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}J(\beta)$$ The last integral can be evaluated by means of integration along a rectangular contour in the complex plane: $$J(\beta)=\frac{\pi}{2}\,\frac{\cos\frac{\beta\pi}{3}}{\cos\frac{\beta\pi}{2}};\quad\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}J(\beta)=\frac{5\pi^3}{72}$$ and $$I(x)=-\frac{\pi}{2}+\frac{4G}{3}\,\frac{1}{x}-\frac{5\pi^3}{144}\,\frac{1}{x^2}+O\Big(\frac{1}{x^3}\Big)$$