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) \[\log (2\sqrt{2})+\frac{\pi }{3}\]
Correct Answer: A
Solution :
\[\int_{0}^{3}{\frac{3x+1}{{{x}^{2}}+9}dx=\frac{3}{2}}\int_{0}^{3}{\frac{2x}{{{x}^{2}}+9}dx+}\int_{0}^{3}{\frac{dx}{{{x}^{2}}+9}}\] \[=\left[ \frac{3}{2}\log ({{x}^{2}}+9)+\frac{1}{3}{{\tan }^{-1}}\left( \frac{x}{3} \right) \right]_{0}^{3}\] \[=\frac{3}{2}(\log 18-\log 9)+\frac{1}{3}\left( \frac{\pi }{4} \right)\] \[=\frac{3}{2}\log 2+\frac{\pi }{12}=\log (2\sqrt{2})+\frac{\pi }{12}\].
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