Explanation

(a)
$\int_{0}^{3}\frac{3x + 1}{x^{2} + 9\ }\ dx = \frac{3}{2}\int_{0}^{3}\frac{2x}{x^{2} + 9\ }\ dx + \int_{0}^{3}\frac{1}{x^{2} + 9\ }\ dx$
$= \frac{3}{2}\left\lbrack \log\left( x^{2} + 9 \right) \right\rbrack_{0}^{3} + \frac{1}{3}\left\lbrack \tan^{- 1}\frac{x}{3} \right\rbrack_{0}^{3}$
$= \frac{3}{2}\left\lbrack \log 18 - \log 9 \right\rbrack + \frac{1}{3}\lbrack\tan^{- 1}(1) - \tan^{- 1}(0)\rbrack$
$= \frac{3}{2}\left\lbrack \log 2 \right\rbrack + \frac{\pi}{12}$
$= \log{(2\sqrt{2})} + \frac{\pi}{12}$