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

The integrand can be recognised as the inverse-hyperbolic sine function, because

$$\sinh^{-1}x=\log_e\!\left(x+\sqrt{x^{2}+1}\right).$$

Hence the integral becomes

$$I=\int_{-1}^{2} \log_e\!\left(x+\sqrt{x^{2}+1}\right)\,dx =\int_{-1}^{2} \sinh^{-1}x \, dx.$$

First we need an antiderivative of $$\sinh^{-1}x.$$
Start with the standard identity

$$\frac{d}{dx}\left(\sinh^{-1}x\right)=\frac{1}{\sqrt{x^{2}+1}}.$$

Use integration by parts with
 • first function $$u=\sinh^{-1}x,$$
 • second function $$dv=dx.$$

This gives

$$\int \sinh^{-1}x\,dx =x\,\sinh^{-1}x-\int x\cdot\frac{1}{\sqrt{x^{2}+1}}\,dx.$$

For the remaining integral, substitute $$t=x^{2}+1 \; (dt=2x\,dx).$$
It simplifies to $$\int \frac{x}{\sqrt{x^{2}+1}}\,dx=\sqrt{x^{2}+1}+C.$$

Therefore

$$\int \sinh^{-1}x\,dx =x\,\sinh^{-1}x-\sqrt{x^{2}+1}+C.$$

Apply the limits −1 to 2:

$$I=\Bigl[x\,\sinh^{-1}x-\sqrt{x^{2}+1}\Bigr]_{-1}^{2}.$$

Evaluate at $$x=2:$$

$$2\,\sinh^{-1}2-\sqrt{2^{2}+1} =2\log_e(2+\sqrt{5})-\sqrt{5}.$$

Evaluate at $$x=-1:$$

$$(-1)\,\sinh^{-1}(-1)-\sqrt{(-1)^{2}+1} =-\,\log_e\!\bigl(-1+\sqrt{2}\bigr)-\sqrt{2}.$$

Subtract the lower value from the upper value:

$$I=\Bigl[2\log_e(2+\sqrt{5})-\sqrt{5}\Bigr] -\Bigl[-\log_e(-1+\sqrt{2})-\sqrt{2}\Bigr].$$

$$I=2\log_e(2+\sqrt{5})-\sqrt{5} +\log_e(\sqrt{2}-1)+\sqrt{2}.$$

Combine the logarithms:

$$I=\sqrt{2}-\sqrt{5} +\log_e\!\Bigl[(2+\sqrt{5})^{2}\,(\sqrt{2}-1)\Bigr].$$

Compute the numerical factor:

$$(2+\sqrt{5})^{2}=4+5+4\sqrt{5}=9+4\sqrt{5},$$
and $$\sqrt{2}-1=\frac{1}{\sqrt{2}+1}.$$

So

$$I=\sqrt{2}-\sqrt{5} +\log_e\!\Bigl[\frac{9+4\sqrt{5}}{1+\sqrt{2}}\Bigr].$$

This matches Option D.

Final answer: $$\sqrt{2}-\sqrt{5}+\log_e\!\left(\dfrac{9+4\sqrt{5}}{1+\sqrt{2}}\right).$$