Evaluate the definite integral $\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$.

(N/A) Let $I = \int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$.
We can split the integral as:
$I = 3 \int_{0}^{2} \frac{2 x}{x^{2}+4} d x + 3 \int_{0}^{2} \frac{1}{x^{2}+4} d x$.
Using the formulas $\int \frac{f'(x)}{f(x)} d x = \log|f(x)|$ and $\int \frac{1}{x^2+a^2} d x = \frac{1}{a} \tan^{-1}(\frac{x}{a})$,we get:
$I = \left[ 3 \log(x^2+4) + \frac{3}{2} \tan^{-1}(\frac{x}{2}) \right]_{0}^{2}$.
Applying the limits:
$I = \left( 3 \log(2^2+4) + \frac{3}{2} \tan^{-1}(\frac{2}{2}) \right) - \left( 3 \log(0^2+4) + \frac{3}{2} \tan^{-1}(\frac{0}{2}) \right)$.
$I = (3 \log 8 + \frac{3}{2} \tan^{-1}(1)) - (3 \log 4 + \frac{3}{2} \tan^{-1}(0))$.
Since $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(0) = 0$:
$I = 3 \log 8 + \frac{3}{2}(\frac{\pi}{4}) - 3 \log 4 - 0$.
$I = 3(\log 8 - \log 4) + \frac{3\pi}{8}$.
$I = 3 \log(\frac{8}{4}) + \frac{3\pi}{8} = 3 \log 2 + \frac{3\pi}{8}$.