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, then $$g(2)$$ is equal to

We are given the integral equation:

$$\int \frac{\log\left(t + \sqrt{1+t^2}\right)}{\sqrt{1+t^2}} dt = \frac{1}{2}(g(t))^2 + c$$

and we need to find $$ g(2) $$.

First, recognize that $$\log\left(t + \sqrt{1+t^2}\right)$$ is the inverse hyperbolic sine function, denoted as $$\sinh^{-1} t$$. So, we rewrite the integral as:

$$\int \frac{\sinh^{-1} t}{\sqrt{1+t^2}} dt$$

To solve this integral, use the substitution $$ u = \sinh^{-1} t $$. Then, $$ t = \sinh u $$, and the derivative is $$ dt = \cosh u du $$.

Also, note that $$ \sqrt{1+t^2} = \sqrt{1 + \sinh^2 u} $$. Using the identity $$ \cosh^2 u - \sinh^2 u = 1 $$, we have $$ 1 + \sinh^2 u = \cosh^2 u $$, so:

$$\sqrt{1+t^2} = \sqrt{\cosh^2 u} = |\cosh u|$$

Since $$\cosh u \geq 1 > 0$$ for all real $$ u $$, we can drop the absolute value: $$\sqrt{1+t^2} = \cosh u$$.

Substitute into the integral:

$$\int \frac{\sinh^{-1} t}{\sqrt{1+t^2}} dt = \int \frac{u}{\cosh u} \cdot \cosh u du$$

The $$\cosh u$$ terms cancel:

$$\int u du$$

Integrate $$ u $$ with respect to $$ u $$:

$$\int u du = \frac{u^2}{2} + c$$

Substitute back $$ u = \sinh^{-1} t $$:

$$\frac{(\sinh^{-1} t)^2}{2} + c$$

Compare this with the given right-hand side $$\frac{1}{2}(g(t))^2 + c$$:

$$\frac{1}{2}(g(t))^2 + c = \frac{(\sinh^{-1} t)^2}{2} + c$$

Thus, $$\frac{1}{2}(g(t))^2 = \frac{(\sinh^{-1} t)^2}{2}$$, which implies:

$$(g(t))^2 = (\sinh^{-1} t)^2$$

Therefore, $$ g(t) = \pm \sinh^{-1} t $$. Since $$\sinh^{-1} t$$ is an increasing function and the integral result is non-negative, we take the positive branch: $$ g(t) = \sinh^{-1} t $$.

Recall that $$\sinh^{-1} t = \log\left(t + \sqrt{1+t^2}\right)$$, so:

$$g(t) = \log\left(t + \sqrt{1+t^2}\right)$$

Now, evaluate $$ g(2) $$:

$$g(2) = \log\left(2 + \sqrt{1+2^2}\right) = \log\left(2 + \sqrt{1+4}\right) = \log\left(2 + \sqrt{5}\right)$$

Compare with the options:

A. $$\frac{1}{\sqrt{5}}\log(2+\sqrt{5})$$

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

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

D. $$\frac{1}{2}\log(2+\sqrt{5})$$

Option C matches $$ g(2) = \log(2 + \sqrt{5}) $$.

Hence, the correct answer is Option C.