Calculate $ \lim_{n\to \infty} \frac{n}{\ln(n)} \int_{1}^{n^2} \frac{\ln(x)}{x^2+nx+n^2}\,dx $ For every $ n\in \mathbb{N} $ and positive $ x $ , $ x \neq 0 $ we consider the function $ f_{n}(x) = \frac{\ln(x)} {x^2+nx+n^2} $
Calculate
$ \lim_{n\to \infty} \frac{n}{\ln(n)} \int_{1}^{n^2} f_{n}(x) \,\mathrm dx$ 
The correct answer should be $ \frac{2\pi \sqrt{3}}{9} $
How to approach this using high-school techniques? The result suggests that  we have to work with an arctangent function probably.

Hint
With $x=ny$ you obtain:
$$n \int_1^{n^2} f_n(x) dx = \int_\frac{1}{n}^n \frac{\ln(ny)}{n^2(y^2+y+1)} n^2 dy=\int_\frac{1}{n}^n \frac{\ln(n)+\ln(y)}{(y^2+y+1)} dy$$
So:
$$\frac{n}{\ln(n)} \int_1^{n^2} f_n(x) dx= \int_\frac{1}{n}^n \frac{1}{(y^2+y+1)} dy+\frac{1}{\ln(n)} \int_\frac{1}{n}^n \frac{\ln(y)}{(y^2+y+1)} dy$$
As $$ \int_0^{+\infty} \frac{|\ln(y)|}{(y^2+y+1)} dy <+ \infty$$ you have:
$$\left|\frac{1}{\ln(n)} \int_\frac{1}{n}^n \frac{\ln(y)}{(y^2+y+1)} dy \right| \leq \frac{1}{\ln(n)} \int_0^{+\infty} \frac{\ln(y)}{(y^2+y+1)} dy \to 0$$
so:
$$\lim_{n \to +\infty} \frac{n}{\ln(n)} \int_1^{n^2} f_n(x) dx = \lim_{n \to + \infty} \int_\frac{1}{n}^n \frac{1}{1+y+y^2} dy =\int_0^{+\infty} \frac{1}{1+y+y^2} dy$$
and the last integral is exactly:
$$\frac{2 \pi \sqrt{3}}{9}$$
