int_{0}^{3} frac{3x + 1}{x^2 + 9} dx = | vedclass

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(A) We evaluate the integral $I = \int_{0}^{3} \frac{3x + 1}{x^2 + 9} dx$. Split the integral into two parts: $I = \frac{3}{2} \int_{0}^{3} \frac{2x}{

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

Vedclass · Answer

(A) We evaluate the integral $I = \int_{0}^{3} \frac{3x + 1}{x^2 + 9} dx$. Split the integral into two parts: $I = \frac{3}{2} \int_{0}^{3} \frac{2x}{x^2 + 9} dx + \int_{0}^{3} \frac{1}{x^2 + 9} dx$. Integrating both parts,we get: $I = \left[ \frac{3}{2} \log(x^2 + 9) + \frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight) ight]_{0}^{3}$. Applying the limits: $I = \left( \frac{3}{2} \log(9 + 9) + \frac{1}{3} 	an^{-1}(1) ight) - \left( \frac{3}{2} \log(0 + 9) + \frac{1}{3} 	an^{-1}(0) ight)$. $I = \frac{3}{2} \log(18) + \frac{1}{3} \left(\frac{\pi}{4}ight) - \frac{3}{2} \log(9) - 0$. $I = \frac{3}{2} \log\left(\frac{18}{9}ight) + \frac{\pi}{12} = \frac{3}{2} \log(2) + \frac{\pi}{12}$. Since $\frac{3}{2} \log(2) = \log(2^{3/2}) = \log(2\sqrt{2})$,the final result is $\log(2\sqrt{2}) + \frac{\pi}{12}$.

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

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