If $$f(x) = \frac{2 - x\cos x}{2 + x\cos x}$$ and $$g(x) = \log_e x$$, then the value of the integral $$\int_{-\pi/4}^{\pi/4} g(f(x)) \, dx$$ is:

We begin by observing that the integrand is built from the two given functions

$$f(x)=\frac{2-x\cos x}{2+x\cos x}\qquad\text{and}\qquad g(x)=\log_e x.$$

Inside the required integral we actually have the composition $$g(f(x))=\log_e\!\bigl(f(x)\bigr),$$ so the integral to be evaluated is

$$I=\int_{-\pi/4}^{\pi/4} \log_e\!\left(\frac{2-x\cos x}{2+x\cos x}\right)\,dx.$$

To simplify the work, we examine the behaviour of the function $$h(x)=\log_e\!\left(\frac{2-x\cos x}{2+x\cos x}\right)$$ under the transformation $$x\mapsto -x$$.

First we compute $$f(-x)=\frac{2-(-x)\cos(-x)}{2+(-x)\cos(-x)}.$$

Because $$\cos(-x)=\cos x$$, the expression becomes

$$f(-x)=\frac{2+x\cos x}{2-x\cos x}.$$

Note that $$f(x)=\frac{2-x\cos x}{2+x\cos x}.$$ Multiplying these two forms gives

$$f(x)\,f(-x)=\left(\frac{2-x\cos x}{2+x\cos x}\right)\!\left(\frac{2+x\cos x}{2-x\cos x}\right)=1.$$

Hence we have the reciprocal relation

$$f(-x)=\frac{1}{f(x)}.$$

Now we apply the natural logarithm to both sides. Using the well-known property $$\log_e\!\left(\frac{1}{u}\right)=-\log_e u,$$ we obtain

$$\log_e\!\bigl(f(-x)\bigr)=-\log_e\!\bigl(f(x)\bigr).$$

Therefore the integrand $$h(x)=\log_e(f(x))$$ is an odd function, which means

$$h(-x)=-h(x).$$

Whenever an odd function is integrated over an interval symmetric about the origin, the positive and negative contributions cancel. Formally, for any odd function $$k(x)$$, the formula

$$\int_{-a}^{a} k(x)\,dx=0$$

holds. Here our limits are $$-a=-\pi/4 \quad\text{and}\quad a=\pi/4,$$ so by direct application we conclude

$$I=\int_{-\pi/4}^{\pi/4} h(x)\,dx=0.$$

The numerical value zero can be written as a natural logarithm because $$\log_e 1=0.$$

So, the value of the given integral is $$\log_e 1.$$

Hence, the correct answer is Option C.