_ ,0 ^ ,3 , 3x + 1 x^2 + 9 dx = - Vidyadip

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

✓

Answer

Correct option: A.

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

a
(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[ {\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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