We are asked to evaluate the indefinite integral $\int -\frac{dx}{2x\sqrt{1-4x^2}}$. We need to find the antiderivative and express the result in terms of inverse hyperbolic secant function ($\text{sech}^{-1}$).

1. Problem Description

We are asked to evaluate the indefinite integral $\int -\frac{dx}{2x\sqrt{1-4x^2}}$. We need to find the antiderivative and express the result in terms of inverse hyperbolic secant function ($\text{sech}^{-1}$).

2. Solution Steps

Let $I = \int -\frac{dx}{2x\sqrt{1-4x^2}}$.

First, let's make a substitution: let $u = 2x$. Then $du = 2dx$, so $dx = \frac{1}{2} du$.

Substituting into the integral:

$I = \int -\frac{\frac{1}{2}du}{2(\frac{1}{2}u)\sqrt{1-u^2}} = \int -\frac{\frac{1}{2}du}{u\sqrt{1-u^2}} = -\frac{1}{2} \int \frac{du}{u\sqrt{1-u^2}}$.

Recall the formula for the integral of $\frac{1}{x\sqrt{a^2-x^2}}$:

$\int \frac{dx}{x\sqrt{a^2 - x^2}} = -\frac{1}{a}\text{sech}^{-1}(\frac{x}{a}) + C$

In our case, we have $\int \frac{du}{u\sqrt{1-u^2}}$, which is equivalent to the above formula with $a=1$.

Therefore, $\int \frac{du}{u\sqrt{1-u^2}} = -\text{sech}^{-1}(u) + C$.

Substituting this back into our original integral:

$I = -\frac{1}{2} \int \frac{du}{u\sqrt{1-u^2}} = -\frac{1}{2} (-\text{sech}^{-1}(u)) + C = \frac{1}{2}\text{sech}^{-1}(u) + C$.

Now, substitute back $u = 2x$:

$I = \frac{1}{2} \text{sech}^{-1}(2x) + C$.

3. Final Answer

The integral evaluates to $\frac{1}{2} \text{sech}^{-1}(2x) + C$. Therefore, the correct answer is A.