If \(\int \frac{\log \left(t+\sqrt{1+\mathrm{t}^2}\right)}{\sqrt{1+\mathrm{t}^2}} \mathrm{dt}=\frac{1}{2}(\mathrm{~g}(\mathrm{t}))^2+\mathrm{c}\)
where c is a constant of integration, then \(g(2)\) is equal to

\(\begin{aligned} & \text { Put } \log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)=y \\ & \Rightarrow {\left[\frac{1}{\mathrm{t}+\sqrt{1+\mathrm{t}^2}}\left(1+\frac{\mathrm{t}}{\sqrt{1+\mathrm{t}^2}}\right)\right] \mathrm{dt}=\mathrm{d} y } \\ & \Rightarrow \frac{1}{\sqrt{1+\mathrm{t}^2}} \mathrm{dt}=\mathrm{d} y \\ & \therefore \quad \int \frac{\log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)}{\sqrt{1+\mathrm{t}^2}} \mathrm{dt}=\int y \mathrm{~d} y \\ &=\frac{y^2}{2}+\mathrm{c}\end{aligned}\)
\(=\frac{\left[\log \left(t+\sqrt{1+\mathrm{t}^2}\right)\right]^2}{2}+\mathrm{c}\)
\(\begin{aligned} & \therefore \quad g(t)=\log \left(t+\sqrt{1+t^2}\right) \\ & \Rightarrow g(2)=\log \left(2+\sqrt{1+2^2}\right)=\log (2+\sqrt{5})\end{aligned}\)