From "Mathematical Analysis" of T.M. Apostol.
\begin{gather*} \lim_{n \rightarrow \infty} \frac{b-a}n \sum_{k=1}^n f \left( a +k\frac{b-a}n \right) =\int_a^bf(x) \ dx \end{gather*}
Use the fact above to compute
\begin{gather*} \lim_{n \rightarrow \infty} \sum_{k=1}^n \frac{n}{k^2+n^2}=\frac{\pi}4 \ \ \ \ \text{and} \ \ \ \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( k^2 + n^2\right)^{-\frac{1}2} = \log\left(1 + \sqrt{2}\right) \end{gather*}
I've tried
\begin{gather*} \lim_{n \rightarrow \infty} \sum_{k=1}^n \frac{n}{k^2+n^2}=\lim_{n \rightarrow \infty} \frac{1}n \sum_{k=1}^n \left( \frac{n}{\sqrt{k^2+n^2}} \right)^2 \end{gather*}
Now $\frac{n}{\sqrt{k^2+n^2}}$ can be interpreted as a sine or a cosine in a right triangle with catheti of lenghts $n$ and $k$. But how exactly the trigonometric functions come into play here?
We have $$\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{k^2+n^2}=\lim_{n\to\infty}\left(\frac{1}{n}\cdot\sum_{k=1}^n \frac{1}{\left(\frac{k}{n}\right)^2+1}\right)=\int_0^1 \frac{1}{x^2+1}\,dx=\arctan x|_0^1=\frac{\pi}{4}\,.$$
For the second one, following the same approach, you will get $$\int_0^1\frac{1}{\sqrt{x^2+1}}\,dx.$$ Evaluate this integral, writing $\mathrm{arsinh}$ in terms of logs to get the desired answer.