∫03 3x+1x2+9dx=∫03 3xx2+9dx+∫03 1x2+9dx =32∫032xx2+9dx+∫031(x)2+(3)2dx =32log(x2+9)03+13tan-1(x3)03 (∵∫f'xfxdx=logfx and∫1a2+x2dx=1atan-1(xa)) =32 log(18) -log9 + 13 tan-1(1)-tan-1(0) =32log(189)+13tan-1(1) =32 log2+13 (π4 ) =log(2)32+π12 =log⁡22+π12

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$\int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} dx =$

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$\ln (2 \sqrt{2}) + \frac{π}{2}$

$\ln (2 \sqrt{2}) + \frac{π}{12}$

$\ln (2 \sqrt{2}) + \frac{π}{3}$

$\ln (2 \sqrt{2}) + \frac{π}{6}$

answer is A.

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Detailed Solution

$\int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} dx = \int_{0}^{3} \frac{3 x}{x^{2} + 9} dx + \int_{0}^{3} \frac{1}{x^{2} + 9} dx = \frac{3}{2} \int_{0}^{3} \frac{2 x}{x^{2} + 9} d x + \int_{0}^{3} \frac{1}{{(x)}^{2} + {(3)}^{2}} d x = \frac{3}{2} {[\log (x^{2} + 9)]}_{0}^{3} + \frac{1}{3} {\tan^{- 1} (\frac{x}{3})}_{0}^{3} (∵ \int \frac{f' (x)}{f (x)} d x = \log |f (x)| a n d \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a})) = \frac{3}{2} [\log (18) - \log 9] + \frac{1}{3} [\tan^{- 1} (1) - \tan^{- 1} (0)] = \frac{3}{2} \log (\frac{18}{9}) + \frac{1}{3} \tan^{- 1} (1) = \frac{3}{2} \log 2 + \frac{1}{3} (\frac{π}{4}) = \log {(2)}^{\frac{3}{2}} + \frac{π}{12} = \log 2 \sqrt{2} + \frac{π}{12}$