1. The problem is to find the integral of the function $$dy = \frac{1}{\frac{1}{2^2} - x}$$ with respect to $x$. 2. Simplify the denominator: $$\frac{1}{2^2} = \frac{1}{4}$$, so the integral becomes $$\int \frac{1}{\frac{1}{4} - x} \, dx = \int \frac{1}{\frac{1}{4} - x} \, dx$$. 3. Rewrite the denominator for clarity: $$\frac{1}{4} - x = \frac{1}{4} - x = -\left(x - \frac{1}{4}\right)$$. 4. Thus, the integral is $$\int \frac{1}{-\left(x - \frac{1}{4}\right)} \, dx = -\int \frac{1}{x - \frac{1}{4}} \, dx$$. 5. The integral of $$\frac{1}{x - a}$$ is $$\ln|x - a| + C$$, therefore: $$-\int \frac{1}{x - \frac{1}{4}} \, dx = -\ln\left|x - \frac{1}{4}\right| + C$$. 6. Hence, the solution to the integral is: $$y = -\ln\left|x - \frac{1}{4}\right| + C$$. Where $C$ is the constant of integration.