Explanation

(a)
$\int_{0}^{3}{\frac{3x + 1}{x^{2} + 9}dx = \frac{3}{2}}\int_{0}^{3}{\frac{2x}{x^{2} + 9}dx + \int_{0}^{3}\frac{dx}{x^{2} + 9}}$
$= \left\lbrack \frac{3}{2}\log{\left( x^{2} + 9 \right) + \frac{1}{3}\tan^{- 1}\left( \frac{x}{3} \right)} \right\rbrack_{0}^{3}$
$= \frac{3}{2}\lbrack(\log{3^{2} + 9) - \log{(0 + 9) + \frac{1}{3}\left\lbrack \tan^{- 1}{\left( \frac{3}{3} \right) - \tan^{- 1}0} \right\rbrack}}$
$= \frac{3}{2}(\log{18 - \log{9) + \frac{1}{3}\left( \frac{\pi}{4} \right)}}$
$= \frac{3}{2}\log{2 + \frac{\pi}{12} = \log{\left( 2\sqrt{2} \right) + \frac{\pi}{12}}}$