int_0^3 frac{3x+1}{x^2+9} dx is equal to : | vedclass

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(A) We need to evaluate the integral $I = \int_0^3 \frac{3x+1}{x^2+9} dx$.Split the integral into two parts:$I = \int_0^3 \frac{3x}{x^2+9} dx + \int_0

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

Vedclass · Answer

(A) We need to evaluate the integral $I = \int_0^3 \frac{3x+1}{x^2+9} dx$.Split the integral into two parts:$I = \int_0^3 \frac{3x}{x^2+9} dx + \int_0^3 \frac{1}{x^2+9} dx$For the first part, let $u = x^2+9$, then $du = 2x dx$, so $x dx = \frac{du}{2}$.$\int_0^3 \frac{3x}{x^2+9} dx = \frac{3}{2} \int_9^{18} \frac{du}{u} = \frac{3}{2} [\log |u|]_9^{18} = \frac{3}{2} (\log 18 - \log 9) = \frac{3}{2} \log 2 = \log (2^{3/2}) = \log (2\sqrt{2})$.For the second part, use the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$:$\int_0^3 \frac{1}{x^2+3^2} dx = [\frac{1}{3} 	an^{-1}(\frac{x}{3})]_0^3 = \frac{1}{3} (	an^{-1}(1) - 	an^{-1}(0)) = \frac{1}{3} (\frac{\pi}{4} - 0) = \frac{\pi}{12}$.Adding both parts, we get $I = \log (2\sqrt{2}) + \frac{\pi}{12}$.

$\int_0^3 \frac{3x+1}{x^2+9} dx$ is equal to :

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