We need to find the indefinite integral of the function $\frac{x+4}{x^2+4}$.

1. Problem Description

We need to find the indefinite integral of the function $\frac{x+4}{x^2+4}$.

2. Solution Steps

We can split the integral into two parts:

$\int \frac{x+4}{x^2+4} dx = \int \frac{x}{x^2+4} dx + \int \frac{4}{x^2+4} dx$

Let $I_1 = \int \frac{x}{x^2+4} dx$ and $I_2 = \int \frac{4}{x^2+4} dx$.

For $I_1$, let $u = x^2 + 4$. Then $du = 2x dx$, so $x dx = \frac{1}{2} du$.

Then $I_1 = \int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C_1 = \frac{1}{2} \ln(x^2+4) + C_1$. (Since $x^2+4$ is always positive, we can drop the absolute value).

For $I_2$, we have $I_2 = \int \frac{4}{x^2+4} dx = 4 \int \frac{1}{x^2+2^2} dx$.

Recall that $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$.

So $I_2 = 4 \cdot \frac{1}{2} \arctan(\frac{x}{2}) + C_2 = 2 \arctan(\frac{x}{2}) + C_2$.

Therefore,

$\int \frac{x+4}{x^2+4} dx = I_1 + I_2 = \frac{1}{2} \ln(x^2+4) + 2 \arctan(\frac{x}{2}) + C$, where $C = C_1 + C_2$.

3. Final Answer

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