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

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(D) 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{2t}{2\sq

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

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(D) Let $y = \log \left(t+\sqrt{1+t^2}\right)$.Then,differentiating with respect to $t$,we get $dy = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{2t}{2\sqrt{1+t^2}}\right) dt$.Simplifying the term inside the parenthesis: $1 + \frac{t}{\sqrt{1+t^2}} = \frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}}$.Thus,$dy = \frac{1}{t+\sqrt{1+t^2}} \cdot \frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}} dt = \frac{1}{\sqrt{1+t^2}} dt$.Substituting this into the integral: $\int y \, dy = \frac{y^2}{2} + c$.Comparing this with the given expression $\frac{1}{2}[g(t)]^2 + c$,we identify $g(t) = y = \log \left(t+\sqrt{1+t^2}\right)$.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

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