$\int \frac{\mathrm{d} x}{3 \cos 2 x+5}$ equals

Let $I = \int \frac{\mathrm{d}x}{3 \cos 2x + 5}$. Using the trigonometric substitution $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$, we transform the integral:

$I = \int \frac{\mathrm{d}x}{3\left(\frac{1 - \tan^2 x}{1 + \tan^2 x}\right) + 5} = \int \frac{1 + \tan^2 x}{8 + 2\tan^2 x}$ $\mathrm{d}x$.

Recognizing that $1 + \tan^2 x = \sec^2 x$, we have $I = \int \frac{\sec^2 x \mathrm{d}x}{8 + 2\tan^2 x}$. With the substitution $t = \tan x$, where $\mathrm{d}t = \sec^2 x \mathrm{d}x$, the integral becomes:

$I = \int \frac{\mathrm{d}t}{8 + 2t^2} = \frac{1}{2} \int \frac{\mathrm{d}t}{4 + t^2}$.

Applying the standard antiderivative formula $\int \frac{\mathrm{d}u}{a^2 + u^2} = \frac{1}{a} \arctan \left(\frac{u}{a}\right) + C$ with $a = 2$, we obtain:

$I = \frac{1}{4} \arctan \left(\frac{t}{2}\right) + C$.

Substituting back $t = \tan x$ gives:

$I = \frac{1}{4} \arctan \left(\frac{\tan x}{2}\right) + C$.

This result corresponds to option C.