Solve the differential equation: 4xydx + (x2 + 1)dy = 0.

The given differential equation is:

$$4xydx + (x^2 + 1)dy = 0$$

Rearrange the terms to separate $x$ and $y$:

$$(x^2 + 1)dy = -4xydx$$

Divide both sides by $y(x^2 + 1)$:

$$\frac{dy}{y} = \frac{-4x}{x^2 + 1}dx$$

$$\int \frac{1}{y} dy = \int \frac{-4x}{x^2 + 1} dx$$

For the integral on the right side, let $u = x^2 + 1$. Then $du = 2x dx$.
So, $-4x dx = -2(2x dx) = -2du$.
The integral becomes:

$$\int \frac{1}{y} dy = \int \frac{-2}{u} du$$

$$\ln|y| = -2\ln|u| + C_1$$

Substitute $u = x^2 + 1$ back into the equation:

$$\ln|y| = -2\ln|x^2 + 1| + C_1$$

Using logarithm properties, $-2\ln|x^2 + 1| = \ln|(x^2 + 1)^{-2}| = \ln\left(\frac{1}{(x^2 + 1)^2}\right)$.

$$\ln|y| = \ln\left(\frac{1}{(x^2 + 1)^2}\right) + C_1$$

Let $C_1 = \ln|C|$, where $C$ is an arbitrary constant.

$$\ln|y| = \ln\left(\frac{1}{(x^2 + 1)^2}\right) + \ln|C|$$

Combine the logarithmic terms:

$$\ln|y| = \ln\left(\frac{C}{(x^2 + 1)^2}\right)$$

Exponentiate both sides:

$$|y| = \frac{C}{(x^2 + 1)^2}$$

Since $C$ is an arbitrary constant, it can absorb the absolute value sign.

$$y = \frac{C}{(x^2 + 1)^2}$$