We will denote
$$s = \sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \dots + \sqrt{1 + \frac{1}{2021^2} + \frac{1}{2022^2}}$$
Notice that
$$1 + \frac{1}{k^2} + \frac{1}{(k+1)^2} = \frac{k^2(k+1)^2 + (k+1)^2 + k^2}{k^2(k+1)^2} = \frac{(k(k+1)+1)^2}{(k(k+1))^2},$$
which yields
$$\sqrt{1 + \frac{1}{k^2} + \frac{1}{(k+1)^2}} = \frac{k(k+1)+1}{k(k+1)} = 1 + \frac{1}{k(k+1)} = 1 + \frac{1}{k} - \frac{1}{k+1}.$$
Therefore
$$$$
s = 2021 + $\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{2021} - \frac{1}{2022}\right)$ = 2021 + $\frac{2021}{2022}.$