If int frac{log left(t+sqrt{1+t^2}right)}{sqr | vedclass

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(B) Let $y = \log \left(t+\sqrt{1+t^2}\right)$.Then,differentiating with respect to $t$,we get $dy = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{t}{\sqrt

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$\log (2+\sqrt{5})$

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(B) Let $y = \log \left(t+\sqrt{1+t^2}\right)$.Then,differentiating with respect to $t$,we get $dy = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{t}{\sqrt{1+t^2}}\right) dt$.Simplifying the expression inside the derivative: $dy = \frac{1}{t+\sqrt{1+t^2}} \left(\frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}}\right) dt = \frac{1}{\sqrt{1+t^2}} dt$.Substituting this into the integral: $\int y \, dy = \frac{y^2}{2} + c$.Thus,$\frac{1}{2} \left[\log \left(t+\sqrt{1+t^2}\right)\right]^2 + c = \frac{1}{2} (g(t))^2 + c$.Comparing both sides,we find $g(t) = \log \left(t+\sqrt{1+t^2}\right)$.Therefore,$g(2) = \log \left(2+\sqrt{1+2^2}\right) = \log (2+\sqrt{5})$.

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

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