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

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