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

Let $$I = \int_{-1}^{1} \log_e\!\left(\sqrt{1-x}+\sqrt{1+x}\right)dx$$. Since the integrand is an even function of $$x$$, we have $$I = 2\int_0^1 \log_e\!\left(\sqrt{1-x}+\sqrt{1+x}\right)dx$$.

Apply integration by parts with $$u = \log_e(\sqrt{1-x}+\sqrt{1+x})$$ and $$dv = dx$$: $$I = 2\Bigl[x\log_e(\sqrt{1-x}+\sqrt{1+x})\Bigr]_0^1 - 2\int_0^1 x \cdot \frac{d}{dx}\!\left[\log_e(\sqrt{1-x}+\sqrt{1+x})\right]dx.$$

The boundary term: at $$x=1$$, value is $$\log_e\sqrt{2} = \frac{1}{2}\log_e 2$$; at $$x=0$$, the term is $$0$$. So the boundary contributes $$2 \cdot \frac{1}{2}\log_e 2 = \log_e 2$$.

For the derivative, using $$\frac{d}{dx}(\sqrt{1-x}) = \frac{-1}{2\sqrt{1-x}}$$ and $$\frac{d}{dx}(\sqrt{1+x}) = \frac{1}{2\sqrt{1+x}}$$: $$\frac{d}{dx}\!\left[\log_e(\sqrt{1-x}+\sqrt{1+x})\right] = \frac{-\frac{1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}}}{\sqrt{1-x}+\sqrt{1+x}} = \frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^2}\,(\sqrt{1-x}+\sqrt{1+x})}.$$

Multiplying numerator and denominator by $$(\sqrt{1-x}-\sqrt{1+x})$$, noting $$(\sqrt{1-x}-\sqrt{1+x})^2 = 2-2\sqrt{1-x^2}$$: $$= \frac{2-2\sqrt{1-x^2}}{2\sqrt{1-x^2}\cdot(-2x)} = \frac{\sqrt{1-x^2}-1}{2x\sqrt{1-x^2}}.$$

Therefore the integral term becomes: $$-2\int_0^1 x \cdot \frac{\sqrt{1-x^2}-1}{2x\sqrt{1-x^2}}\,dx = -\int_0^1 \frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}}\,dx = -\int_0^1\!\left(1 - \frac{1}{\sqrt{1-x^2}}\right)dx.$$

Evaluating: $$-\left[x - \sin^{-1}x\right]_0^1 = -\!\left[\left(1 - \frac{\pi}{2}\right) - 0\right] = \frac{\pi}{2} - 1.$$

Hence $$I = \log_e 2 + \dfrac{\pi}{2} - 1$$.