Limits of sums as Riemann sums

Principle. For $f$ continuous on $[0,1]$, the Riemann sums $\frac1n\sum_{k=1}^nf\!\left(\frac kn\right)$ converge to $\int_0^1f$.

$A$. Dividing numerator and denominator by $n^2$,
$$\frac n{n^2+k^2}=\frac1n\cdot\frac1{1+(k/n)^2}.$$
So $A=\int_0^1\frac{dx}{1+x^2}=\arctan1=\dfrac\pi4$.

$B$. Similarly $\dfrac1{n+k}=\dfrac1n\cdot\dfrac1{1+k/n}$, so
$$B=\int_0^1\frac{dx}{1+x}=\ln2.$$

Numerical check. For $n=100$: the first sum is $0.7829$ (close to $\frac\pi4=0.7854$) and the second is $0.6907$ (close to $\ln2=0.6931$).

Remark. This is the standard way to evaluate limits of sums: spot the $\frac1n$ and the ratio $\frac kn$.