$\int_0^2 \frac{3 x+1}{x^2+4} d x=$

The integral $I = \int_0^2 \frac{3x+1}{x^2+4} dx$ separates into two components:

$I = \int_0^2 \frac{3x}{x^2+4} dx + \int_0^2 \frac{1}{x^2+4} dx$

For the first term, substitute $u = x^2 + 4$ so that $du = 2x dx$ and $x dx = \frac{1}{2} du$:
When $x = 0$, $u = 4$
When $x = 2$, $u = 8$
This yields $\frac{3}{2} \int_4^8 \frac{1}{u} du = \frac{3}{2} \ln\left|u\right| \Big|_4^8 = \frac{3}{2} (\ln 8 - \ln 4) = \frac{3}{2} \ln 2 = \ln(2\sqrt{2})$

The second term follows the standard form:
$\int_0^2 \frac{1}{x^2+4} dx = \frac{1}{2} \arctan\left(\frac{x}{2}\right) \Big|_0^2 = \frac{1}{2}\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{8}$

The complete integral evaluates to $\ln(2\sqrt{2}) + \frac{\pi}{8}$, matching option C.