Proving $\lim\limits_{n{\rightarrow}\infty} \int_{0}^{n} \frac{{\sqrt{x}}\ln x}{1+x^2}dx=\pi$ 
Show that $$\lim\limits_{n{\rightarrow}\infty} \int_{0}^{n} \frac{x^{\frac{1}{2}}\ln x}{1+x^2}dx=\pi$$

After a number of transformations I ended having $$4\int_{0}^{\frac{1}{n^2}}\frac{\ln(x)}{x^4+1}dx$$ From here on I have
 no idea on how to continue.

Here is an approach that will make use of the derivative of the Beta function $\operatorname{B} (m,n)$ of the form 
$$\operatorname{B} (m,n) = \int_0^\infty \frac{x^{m - 1}}{(1 + x)^{m + n}} \, dx, \quad m,n > 0.$$
Let
$$I = \int_0^\infty \frac{\sqrt{x} \, \ln x}{1 + x^2} \, dx.$$
Enforcing a substitution of $x \mapsto \sqrt{x}$ leads to
$$I = \frac{1}{4} \int_0^\infty \frac{x^{-1/4} \ln x}{1 + x} \, dx.$$
Now consider 
$$J(s) = \frac{1}{4} \int_0^\infty \frac{x^s}{1 + x} \, dx, \qquad - 1 < s < 0.$$
On differentiating with respect to $s$ we have
$$J'(s) = \frac{1}{4} \int_0^\infty \frac{x^s \ln x}{1 + x} \, dx.$$
For our integral we see that $I = J'(-1/4)$.
Now
\begin{align}
J(s) &= \frac{1}{4} \int_0^\infty \frac{x^{(s + 1) - 1}}{(1 + x)^{(s + 1) + (-s)}} \, dx\\
&= \frac{1}{4} \operatorname{B} (s + 1, -s)\\
&= \frac{1}{4} \Gamma (s + 1) \Gamma (-s).
\end{align}
From the reflexion formula for the Gamma function, namely
$$\Gamma (1 - x) \Gamma (x) = \frac{\pi}{\sin (\pi x)},$$
replacing $x$ with $-s$ gives
$$\Gamma (s + 1) \Gamma (-s) = - \pi \operatorname{cosec} (\pi s).$$
Thus
$$J(s) = -\frac{\pi}{4} \operatorname{cosec} (\pi s).$$
Differentiating with respect to $s$ gives
$$J'(s) = \frac{\pi^2}{4} \cot (\pi s) \operatorname{cosec} (\pi s).$$
Setting $s = -1/4$ gives
$$J \left (-\frac{1}{4} \right ) = \frac{\pi^2}{4} \cot \left (-\frac{\pi}{4} \right ) \operatorname{cosec} \left (-\frac{\pi}{4} \right ),$$
or
$$\int_0^\infty \frac{\sqrt{x} \, \ln x}{1 + x^2} \, dx = \frac{\pi^2}{2 \sqrt{2}}.$$ 
