We are given a first-order differential equation $(x^2 + 4)y' + xy^2 + x = 0$ with the initial condition $y(1) = 3$. We need to find the solution to this initial value problem.

1. Problem Description

We are given a first-order differential equation $(x^2 + 4)y' + xy^2 + x = 0$ with the initial condition $y(1) = 3$. We need to find the solution to this initial value problem.

2. Solution Steps

First, rewrite the differential equation as:

$(x^2 + 4)\frac{dy}{dx} + xy^2 + x = 0$

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

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

Now, integrate both sides:

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

The left integral is $\arctan(y)$. For the right integral, let $u = x^2 + 4$, so $du = 2x dx$, and $x dx = \frac{1}{2} du$.

$\int -\frac{x}{x^2 + 4} dx = -\int \frac{1}{u} \frac{1}{2} du = -\frac{1}{2} \int \frac{1}{u} du = -\frac{1}{2} \ln|u| + C = -\frac{1}{2} \ln(x^2 + 4) + C$

So, we have:

$\arctan(y) = -\frac{1}{2} \ln(x^2 + 4) + C$

Now, apply the initial condition $y(1) = 3$:

$\arctan(3) = -\frac{1}{2} \ln(1^2 + 4) + C$

$\arctan(3) = -\frac{1}{2} \ln(5) + C$

$C = \arctan(3) + \frac{1}{2} \ln(5)$

So the solution is:

$\arctan(y) = -\frac{1}{2} \ln(x^2 + 4) + \arctan(3) + \frac{1}{2} \ln(5)$

$\arctan(y) = \arctan(3) + \frac{1}{2} \ln(5) - \frac{1}{2} \ln(x^2 + 4)$

$\arctan(y) = \arctan(3) + \frac{1}{2} \ln(\frac{5}{x^2 + 4})$

$y = \tan\left( \arctan(3) + \frac{1}{2} \ln\left(\frac{5}{x^2 + 4}\right) \right)$

3. Final Answer

$y = \tan\left( \arctan(3) + \frac{1}{2} \ln\left(\frac{5}{x^2 + 4}\right) \right)$