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

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Aicharya · Accepted Answer

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

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

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