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

Let $I = \int \log(t+\sqrt{1+t^2})\sqrt{1+t^2}dt$.
Let $t = \sinh u$. Then $dt = \cosh u du$.
Also, $\sqrt{1+t^2} = \sqrt{1+\sinh^2 u} = \cosh u$.
Then
$I = \int \log(\sinh u + \cosh u)\cosh^2 u du$
Since $\sinh u + \cosh u = e^u$,
$I = \int \log(e^u)\cosh^2 u du = \int u \cosh^2 u du$
$\cosh^2 u = \frac{1+\cosh 2u}{2}$
$I = \int u \frac{1+\cosh 2u}{2} du = \frac{1}{2} \int (u + u\cosh 2u) du$
$= \frac{1}{2} \left[ \frac{u^2}{2} + \frac{u\sinh 2u}{2} - \frac{\cosh 2u}{4} \right] + C$
$= \frac{1}{2} \left[ \frac{u^2}{2} + \frac{u(2\sinh u \cosh u)}{2} - \frac{\cosh^2 u + \sinh^2 u}{2} \right] + C$
$= \frac{1}{4} \left[ u^2 + 2u\sinh u \cosh u - (\cosh^2 u + \sinh^2 u) \right] + C$
Since $t = \sinh u$, $u = \sinh^{-1} t = \log(t+\sqrt{1+t^2})$ and $\cosh u = \sqrt{1+t^2}$
$I = \frac{1}{4} \left[ (\log(t+\sqrt{1+t^2}))^2 + 2t\sqrt{1+t^2}\log(t+\sqrt{1+t^2}) - (1+2t^2) \right] + C$
Comparing this with $\frac{1}{2}(g(t))^2 + C$,
$\frac{1}{2}(g(t))^2 = \frac{1}{4} \left[ (\log(t+\sqrt{1+t^2}))^2 + 2t\sqrt{1+t^2}\log(t+\sqrt{1+t^2}) - (1+2t^2) \right]$
When $t=2$, $g(2) = \sqrt{\frac{1}{2} \left[ (\log(2+\sqrt{5}))^2 + 4\sqrt{5}\log(2+\sqrt{5}) - 9 \right]}$
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