If int_{1}^{2} frac{dx}{(x^2 - 2x + 4)^{3/2}} | vedclass

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(A) Let $I = \int_{1}^{2} \frac{dx}{((x-1)^2 + 3)^{3/2}}$.Substitute $x-1 = \sqrt{3} 	an 	heta$,so $dx = \sqrt{3} \sec^2 	heta \, d	heta$.When $x=

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(A) Let $I = \int_{1}^{2} \frac{dx}{((x-1)^2 + 3)^{3/2}}$.Substitute $x-1 = \sqrt{3} 	an 	heta$,so $dx = \sqrt{3} \sec^2 	heta \, d	heta$.When $x=1$,$	an 	heta = 0 \implies 	heta = 0$.When $x=2$,$	an 	heta = \frac{1}{\sqrt{3}} \implies 	heta = \frac{\pi}{6}$.$I = \int_{0}^{\pi/6} \frac{\sqrt{3} \sec^2 	heta \, d	heta}{(\sqrt{3} \sec 	heta)^3} = \int_{0}^{\pi/6} \frac{\sqrt{3} \sec^2 	heta}{3\sqrt{3} \sec^3 	heta} \, d	heta$.$I = \frac{1}{3} \int_{0}^{\pi/6} \cos 	heta \, d	heta = \frac{1}{3} [\sin 	heta]_{0}^{\pi/6} = \frac{1}{3} 	imes \frac{1}{2} = \frac{1}{6}$.Given $\frac{k}{k+5} = \frac{1}{6}$,we have $6k = k+5$,which implies $5k = 5$,so $k = 1$.

If $\int_{1}^{2} \frac{dx}{(x^2 - 2x + 4)^{3/2}} = \frac{k}{k+5}$,then $k$ is equal to

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