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

Let us denote the required integral by

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

First we analyse the behaviour of the integrand. For every real number $$x$$ we have the algebraic identity

$$\bigl(x+\sqrt{x^{2}+1}\bigr)\bigl(\sqrt{x^{2}+1}-x\bigr)=\bigl(\sqrt{x^{2}+1}\bigr)^{2}-x^{2}=x^{2}+1-x^{2}=1.$$

Taking natural logarithms on both sides gives

$$\log\!\left(x+\sqrt{x^{2}+1}\right)+\log\!\left(\sqrt{x^{2}+1}-x\right)=\log 1=0.$$

So we can write

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

Next we replace $$x$$ by $$-x$$ inside the original logarithm:

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

But $$-x+\sqrt{x^{2}+1}=\sqrt{x^{2}+1}-x,$$ so using the previous relation we obtain

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

Hence the integrand is an odd function:

$$f(-x)=-f(x)\quad\text{where}\quad f(x)=\log\!\left(x+\sqrt{x^{2}+1}\right).$$

The standard result for definite integrals of odd functions tells us that

$$\int_{-a}^{a}f(x)\,dx=0\quad\text{whenever}\quad f(-x)=-f(x).$$

Here the limits are symmetric (from $$-1$$ to $$1$$), so the entire area cancels out:

$$I=\int_{-1}^{1}\log\!\left(x+\sqrt{x^{2}+1}\right)\,dx=0.$$

Hence, the correct answer is Option B.