int_0^2 frac{3 x+1}{x^2+4} d x= | vedclass

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(C) We need to evaluate the integral $I = \int_0^2 \frac{3x+1}{x^2+4} dx$. Split the integral into two parts: $I = \int_0^2 \frac{3x}{x^2+4} dx + \int

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$\frac{3}{2} \log 2 + \frac{\pi}{8}$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int_0^2 \frac{3x+1}{x^2+4} dx$. Split the integral into two parts: $I = \int_0^2 \frac{3x}{x^2+4} dx + \int_0^2 \frac{1}{x^2+4} dx$. For the first part,let $u = x^2+4$,then $du = 2x dx$,so $x dx = \frac{du}{2}$. $\int_0^2 \frac{3x}{x^2+4} dx = \frac{3}{2} \int_4^8 \frac{du}{u} = \frac{3}{2} [\log |u|]_4^8 = \frac{3}{2} (\log 8 - \log 4) = \frac{3}{2} \log(\frac{8}{4}) = \frac{3}{2} \log 2$. For the second part,use the standard integral $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$. $\int_0^2 \frac{1}{x^2+2^2} dx = [\frac{1}{2} 	an^{-1}(\frac{x}{2})]_0^2 = \frac{1}{2} 	an^{-1}(1) - \frac{1}{2} 	an^{-1}(0) = \frac{1}{2} (\frac{\pi}{4}) - 0 = \frac{\pi}{8}$. Combining both parts,$I = \frac{3}{2} \log 2 + \frac{\pi}{8}$.

$\int_0^2 \frac{3 x+1}{x^2+4} d x=$

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