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

Question

Aicharya · Accepted Answer

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

If $\int \frac{\log \left(t+\sqrt{1+\mathrm{t}^2}\right)}{\sqrt{1+\mathrm{t}^2}}…

Solution