Complete the square in the denominator.

The denominator is $x^2 + 8x + 25$. To complete the square for $x^2 + bx$, we add and subtract $(\frac{b}{2})^2$. Here, $b=8$, so $(\frac{8}{2})^2 = 4^2 = 16$.

$$x^2 + 8x + 25 = x^2 + 8x + 16 - 16 + 25$$ $$x^2 + 8x + 25 = (x+4)^2 + 9$$

Let $u = x+4$. Then, $du = dx$.
The integral becomes:

$$\int \frac{1}{u^2 + 3^2} du$$

This matches the standard integral form $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$, where $a=3$.

$$\frac{1}{3} \arctan\left(\frac{u}{3}\right) + C$$

Substitute $u = x+4$ back into the expression:

$$\frac{1}{3} \arctan\left(\frac{x+4}{3}\right) + C$$