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

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(D) Let $I = \int \frac{\log (t+\sqrt{1+t^{2}})}{\sqrt{1+t^{2}}} dt$. Substitute $u = \log (t+\sqrt{1+t^{2}})$. Then,$du = \frac{1}{t+\sqrt{1+t^{2}}}

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

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(D) Let $I = \int \frac{\log (t+\sqrt{1+t^{2}})}{\sqrt{1+t^{2}}} dt$. Substitute $u = \log (t+\sqrt{1+t^{2}})$. Then,$du = \frac{1}{t+\sqrt{1+t^{2}}} \cdot \left( 1 + \frac{2t}{2\sqrt{1+t^{2}}} \right) dt = \frac{1}{t+\sqrt{1+t^{2}}} \cdot \left( \frac{\sqrt{1+t^{2}}+t}{\sqrt{1+t^{2}}} \right) dt = \frac{1}{\sqrt{1+t^{2}}} dt$. Therefore,$I = \int u du = \frac{u^{2}}{2} + C$. Comparing this with the given expression $I = \frac{1}{2}[g(t)]^{2} + C$,we get $g(t) = u = \log (t+\sqrt{1+t^{2}})$. Now,evaluating at $t = 2$: $g(2) = \log (2+\sqrt{1+2^{2}}) = \log (2+\sqrt{5})$.

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

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