$\lim_{n\rightarrow \infty} n^{2} \sum_{k=1}^{n}\frac{1}{(n^2+k^2)(\sqrt{n^2+k^2}+n)}$. Find $$\lim_{n\rightarrow \infty} n^{2}\sum_{k=1}^{n}\frac{1}{(n^2+k^2)(\sqrt{n^2+k^2}+n)}.$$
My approach:
\begin{align*}
I& =\lim_{n\rightarrow \infty} n^{2}\sum_{k=1}^{n}\frac{1}{(n^2+k^2)(\sqrt{n^2+k^2}+n)}=\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{(1+(\frac{k}{n})^2)(1+\sqrt{1+(\frac{k}{n})^2})}\\ & =\int_{0}^{1} \frac{1}{(1+x^2)(1+\sqrt{1+x^2})}dx \Longrightarrow 
\end{align*}
\begin{align*}I & =\int_{0}^{1} \left(\frac{1}{x^2+1}-\frac{1}{\sqrt{x^2+1}}+\frac{1}{1+\sqrt{x^2+1}}\right)dx\\ & =\arctan(1)-\ln(1+\sqrt{2})+\int_{0}^{1}\frac{1}{1+\sqrt{x^2+1}}dx\end{align*} and at this point I am stuck.
Any help, please?

For the last integral consider the indefinite integral
$$I=\int \frac{1}{\sqrt{x^2+1}+1}dx=\int \frac{\sqrt{x^2+1}-1}{x^2}dx$$
Here,  you can use the hyperbolic trig substitution $x=\sinh t$ for the first summand in the numerator which yields
$$\int \frac{\sqrt{x^2+1}}{x^2}dx=\int \coth ^2 t ~ dt=t-\coth t+C$$
which after substituting back yields an antiderivative for $I$
$$I=\sinh ^{-1}x-\frac{\sqrt{x^2+1}-1}{x}+C$$
Note that this function is, as expected, regular at $x=0$, unlike the one in the 2nd equation above.
