The value of the integral $$\int_{-1}^{2} \log_e\left(x + \sqrt{x^2 + 1}\right) dx$$ is

Let $$f(x) = \log_e (x + \sqrt{x^2 + 1})$$. This is an odd function ($$f(-x) = -f(x)$$).
Integration by parts: $$\int 1 \cdot f(x) \, dx = x f(x) - \int \frac{x}{\sqrt{x^2+1}} \, dx = x f(x) - \sqrt{x^2+1}$$.
Apply limits:
$$[x \log(x + \sqrt{x^2+1}) - \sqrt{x^2+1}]_{-1}^{2}$$
$$= (2 \log(2 + \sqrt{5}) - \sqrt{5}) - (-1 \log(-1 + \sqrt{2}) - \sqrt{2})$$
$$= \log(2+\sqrt{5})^2 - \sqrt{5} + \log(\sqrt{2}-1) + \sqrt{2}$$
$$= \log(9+4\sqrt{5}) - \sqrt{5} - \log(\sqrt{2}+1) + \sqrt{2}$$
(Note: $$\sqrt{2}-1 = 1/(\sqrt{2}+1)$$)
$$= \sqrt{2} - \sqrt{5} + \log_e \left( \frac{9+4\sqrt{5}}{1+\sqrt{2}} \right)$$
Correct Option: D