The problem asks us to evaluate the integral $\int -\frac{dx}{2x\sqrt{1-4x^2}}$.

1. Problem Description

The problem asks us to evaluate the integral $\int -\frac{dx}{2x\sqrt{1-4x^2}}$.

2. Solution Steps

We want to evaluate the integral

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

Let $u = 2x$, so $du = 2dx$, and $dx = \frac{1}{2} du$. The integral becomes

$$-\frac{1}{2} \int \frac{\frac{1}{2} du}{\frac{1}{2} u \sqrt{1-u^2}} = -\frac{1}{2} \int \frac{du}{u\sqrt{1-u^2}}$$

We know that

$$\int \frac{du}{u\sqrt{a^2-u^2}} = -\frac{1}{a} \ln \left| \frac{a + \sqrt{a^2-u^2}}{u} \right| + C$$

In our case, $a=1$, so we have

$$\int \frac{du}{u\sqrt{1-u^2}} = -\ln \left| \frac{1 + \sqrt{1-u^2}}{u} \right| + C$$

Thus,

$$-\frac{1}{2} \int \frac{du}{u\sqrt{1-u^2}} = -\frac{1}{2} \left( -\ln \left| \frac{1 + \sqrt{1-u^2}}{u} \right| \right) + C = \frac{1}{2} \ln \left| \frac{1 + \sqrt{1-u^2}}{u} \right| + C$$

Substitute back $u = 2x$, so we get

$$\frac{1}{2} \ln \left| \frac{1 + \sqrt{1-4x^2}}{2x} \right| + C$$

We can use the property of logarithms to write this as

$$\ln \sqrt{\left| \frac{1 + \sqrt{1-4x^2}}{2x} \right|} + C$$

Another approach:

Let $2x = \sin \theta$, so $x = \frac{1}{2} \sin \theta$, and $dx = \frac{1}{2} \cos \theta \, d\theta$. Then $\sqrt{1-4x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta$. The integral becomes

$$\int -\frac{dx}{2x\sqrt{1-4x^2}} = \int -\frac{\frac{1}{2} \cos \theta \, d\theta}{\sin \theta \cos \theta} = -\frac{1}{2} \int \frac{d\theta}{\sin \theta} = -\frac{1}{2} \int \csc \theta \, d\theta$$

We know that $\int \csc \theta \, d\theta = -\ln|\csc \theta + \cot \theta| + C$, so

$$-\frac{1}{2} \int \csc \theta \, d\theta = \frac{1}{2} \ln|\csc \theta + \cot \theta| + C$$

Since $\sin \theta = 2x$, we have $\csc \theta = \frac{1}{2x}$. Also, $\cot \theta = \frac{\sqrt{1-\sin^2 \theta}}{\sin \theta} = \frac{\sqrt{1-4x^2}}{2x}$. Thus,

$$\frac{1}{2} \ln|\csc \theta + \cot \theta| + C = \frac{1}{2} \ln\left|\frac{1}{2x} + \frac{\sqrt{1-4x^2}}{2x}\right| + C = \frac{1}{2} \ln\left|\frac{1 + \sqrt{1-4x^2}}{2x}\right| + C$$

3. Final Answer

$\frac{1}{2} \ln\left|\frac{1 + \sqrt{1-4x^2}}{2x}\right| + C$