The correct answer is $\boxed{\frac{\pi}{4}}$.
We can evaluate the integral by substituting $u = \tan x$, which gives $du = \sec^2 x \, dx$. Substituting gives us:
$$\int_0^{\frac{\pi}{4}} \frac{(1 – \tan x)}{(1 + \tan x)} \, dx = \int_0^1 \frac{(1 – u)}{(1 + u)} \sec^2 x \, du$$
We can now evaluate this integral using partial fractions:
$$\int_0^1 \frac{(1 – u)}{(1 + u)} \sec^2 x \, du = \int_0^1 \left( \frac{1}{2} – \frac{1}{1 + u} \right) \sec^2 x \, du$$
The first integral is easy to evaluate:
$$\int_0^1 \frac{1}{2} \sec^2 x \, du = \frac{\sec^2 x}{2} \bigg|_0^1 = \frac{1}{2}$$
The second integral can be evaluated using the substitution $v = 1 + u$, which gives $du = -dv$. Substituting gives us:
$$\int_0^1 \frac{-1}{v} \sec^2 x \, du = -\int_1^2 \frac{1}{v} \sec^2 x \, dv$$
We can now evaluate this integral using the following identity:
$$\int \frac{\sec^2 x}{v} \, dv = \ln |v| + \tan x + C$$
where $C$ is an arbitrary constant. Substituting gives us:
$$\int_0^1 \frac{-1}{v} \sec^2 x \, dv = -\ln |v| – \tan x \bigg|_1^2 = -\ln 2 – \tan x + \ln 2 + \tan x = -\tan x$$
Therefore, the integral we are looking for is:
$$\int_0^{\frac{\pi}{4}} \frac{(1 – \tan x)}{(1 + \tan x)} \, dx = \frac{1}{2} – \left( -\tan x \right) = \frac{\pi}{4}$$