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$\int \frac{\log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)}{\sqrt{1+\mathrm{t}^2}} \mathrm{dt}=\frac{1}{2}[\mathrm{~g}(\mathrm{t})]^2+\mathrm{c}$, (where $\mathrm{c}$ is a constant of integration) then $g(2)$ is
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The correct answer is:
$\log (2+\sqrt{5})$
Put $\log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)=y$
$\begin{aligned} & \Rightarrow\left[\frac{1}{\mathrm{t}+\sqrt{1+\mathrm{t}^2}}\left(1+\frac{\mathrm{t}}{\sqrt{1+\mathrm{t}^2}}\right)\right] \mathrm{dt}=\mathrm{d} y \\ & \Rightarrow \frac{1}{\sqrt{1+\mathrm{t}^2}} \mathrm{dt}=\mathrm{d} y\end{aligned}$
$\begin{aligned} \therefore \quad \int \frac{\log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)}{\sqrt{1+\mathrm{t}^2}} \mathrm{dt} & =\int y \mathrm{~d} y \\ & =\frac{y^2}{2}+\mathrm{c} \\ & =\frac{\left[\log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)\right]^2}{2}+\mathrm{c}\end{aligned}$
$\begin{aligned} \therefore \quad & g(t)=\log \left(t+\sqrt{1+t^2}\right) \\ \Rightarrow & g(2)=\log \left(2+\sqrt{1+2^2}\right)=\log (2+\sqrt{5})\end{aligned}$
$\begin{aligned} & \Rightarrow\left[\frac{1}{\mathrm{t}+\sqrt{1+\mathrm{t}^2}}\left(1+\frac{\mathrm{t}}{\sqrt{1+\mathrm{t}^2}}\right)\right] \mathrm{dt}=\mathrm{d} y \\ & \Rightarrow \frac{1}{\sqrt{1+\mathrm{t}^2}} \mathrm{dt}=\mathrm{d} y\end{aligned}$
$\begin{aligned} \therefore \quad \int \frac{\log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)}{\sqrt{1+\mathrm{t}^2}} \mathrm{dt} & =\int y \mathrm{~d} y \\ & =\frac{y^2}{2}+\mathrm{c} \\ & =\frac{\left[\log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)\right]^2}{2}+\mathrm{c}\end{aligned}$
$\begin{aligned} \therefore \quad & g(t)=\log \left(t+\sqrt{1+t^2}\right) \\ \Rightarrow & g(2)=\log \left(2+\sqrt{1+2^2}\right)=\log (2+\sqrt{5})\end{aligned}$
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