If int_1^2 frac{dx}{(x^2-2x+4)^{frac{3}{2}}} | vedclass

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(A) Let $I = \int_1^2 \frac{dx}{(x^2-2x+4)^{\frac{3}{2}}} = \int_1^2 \frac{dx}{((x-1)^2+3)^{\frac{3}{2}}}$.Substitute $x-1 = \sqrt{3} 	an 	heta$,so

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(A) Let $I = \int_1^2 \frac{dx}{(x^2-2x+4)^{\frac{3}{2}}} = \int_1^2 \frac{dx}{((x-1)^2+3)^{\frac{3}{2}}}$.Substitute $x-1 = \sqrt{3} 	an 	heta$,so $dx = \sqrt{3} \sec^2 	heta \ d	heta$.When $x=1$,$	heta=0$. When $x=2$,$	an 	heta = \frac{1}{\sqrt{3}}$,so $	heta = \frac{\pi}{6}$.$I = \int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec^2 	heta}{(3 \sec^2 	heta)^{\frac{3}{2}}} \ d	heta = \int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec^2 	heta}{3\sqrt{3} \sec^3 	heta} \ d	heta = \frac{1}{3} \int_0^{\frac{\pi}{6}} \cos 	heta \ d	heta$.$I = \frac{1}{3} [\sin 	heta]_0^{\frac{\pi}{6}} = \frac{1}{3} (\sin \frac{\pi}{6} - \sin 0) = \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 $\int_1^2 \frac{dx}{(x^2-2x+4)^{\frac{3}{2}}} = \frac{k}{k+5}$,then $k$ has the value

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