$\int_0^3 \frac{3 x+1}{x^2+9} d x$ is equal to :

• Correct answer
  A

  $\log (2 \sqrt{2})+\frac{\pi}{12}$
• B

  $\log (2 \sqrt{2})+\frac{\pi}{2}$
• C

  $\log (2 \sqrt{2})+\frac{\pi}{6}$
• D

  $\log \left(2(\sqrt{2})+\frac{\pi}{3}\right.$

Solution

$=\frac{3}{2}\left[\log \left(x^2+9\right)\right]_0^3+\frac{1}{3}\left[\tan ^{-1} \frac{x}{3}\right]_0^3$ $=\frac{3}{2}[\log 18-\log 9]+\frac{1}{3}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right]$ $=\frac{3}{2}[\log 2]+\frac{\pi}{12}$ $=\log (2 \sqrt{2})+\frac{\pi}{12}$

Hence, the correct answer is option: 'A'.