Let f : R $ \to $R be a continuously differentiable function such that f(2) = 6 and f'(2) = ${1 \over {48}}$. If $\int\limits_6^{f\left( x \right)} {4{t^3}} dt$ = (x - 2)g(x), then $\mathop {\lim }\limits_{x \to 2} g\left( x \right)$ is equal to :
The value of $\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x$ is