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

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(A) माना $I = \int_{1}^{2} \frac{dx}{((x-1)^2 + 3)^{3/2}}$.$x-1 = \sqrt{3} 	an 	heta$ प्रतिस्थापन करने पर,$dx = \sqrt{3} \sec^2 	heta \, d	heta$ प

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$1$

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(A) माना $I = \int_{1}^{2} \frac{dx}{((x-1)^2 + 3)^{3/2}}$.$x-1 = \sqrt{3} 	an 	heta$ प्रतिस्थापन करने पर,$dx = \sqrt{3} \sec^2 	heta \, d	heta$ प्राप्त होता है।जब $x=1$,तब $	an 	heta = 0 \implies 	heta = 0$.जब $x=2$,तब $	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}$.दिया गया है कि $\frac{k}{k+5} = \frac{1}{6}$,इसलिए $6k = k+5$,जिसका अर्थ है कि $5k = 5$,अर्थात $k = 1$।

यदि $\int_{1}^{2} \frac{dx}{(x^2 - 2x + 4)^{3/2}} = \frac{k}{k+5}$ है,तो $k$ का मान ज्ञात कीजिए।

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