Evaluate $\lim_\limits{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$ $$\lim_{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$$
My attempt
\begin{align*}
&=\exp \lim_\limits{x \to +\infty} x^2\arctan x \cdot\ln\left[x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right]\\
&=\exp \lim_{x \to +\infty} \frac{\pi}{2}x^2 \ln\left[\frac{\ln x}{\frac{1}{x}} + \frac{\ln \left(1+\frac{1}{x}\right)}{\frac{1}{x}}-\frac{\ln x}{\frac{1}{x}} + \arctan\frac{1}{2x}\right]\\
&=\exp \lim_{x \to +\infty} \frac{\pi}{2}x^2\ln\left[1 + \arctan \frac{1}{2x}\right]
\end{align*}
Here I don't know what to do. One thing I did was to multiply and divide by $\arctan \frac{1}{2x}$ in order to get rid of the logarithm. But I ended up with 
$$\frac{\pi}{2}x^2\arctan \frac{1}{2x}$$
$\arctan \frac{1}{2x} \to \pi/2$ as $x \to +\infty$ but that doesn't lead to the right answer (I'd get a limit of $+\infty$).

Let $f(x)$ be your function.
we have
$x\ln(1+\frac{1}{x})+\arctan(\frac{1}{2x})=$
$=x(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3})+(\frac{1}{2x}-\frac{1}{24x^3})+\frac{1}{x^3}\epsilon(x)$
$=1+\frac{1}{3x^2}(1+\epsilon(x))$
thus
$\ln(f(x))\sim \frac{1}{3}\arctan(x)\;\; (x\to+\infty)$
and
$$\lim_{x\to+\infty}f(x)=e^{\frac{\pi}{6}}.$$
