MCQ
$\int_{\,0}^{\,3} {\,\frac{{3x + 1}}{{{x^2} + 9}}dx = } $
- ✓$\log (2\sqrt 2 ) + \frac{\pi }{{12}}$
- B$\log (2\sqrt 2 ) + \frac{\pi }{2}$
- C$\log (2\sqrt 2 ) + \frac{\pi }{6}$
- D$\log (2\sqrt 2 ) + \frac{\pi }{3}$
$ = \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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