int frac{dx}{x^2 sqrt{4+x^2}} ની કિંમત શોધો. | vedclass

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(C) ધારો કે $I = \int \frac{dx}{x^2 \sqrt{4+x^2}}$.$x = 2 	an 	heta$ આદેશ લેતા,$dx = 2 \sec^2 	heta \ d	heta$ મળે.સંકલન $I = \int \frac{2 \sec^2 \

Vedclass · Accepted Answer

$\frac{-1}{4x} \sqrt{4+x^2}+C$

Vedclass · Answer

(C) ધારો કે $I = \int \frac{dx}{x^2 \sqrt{4+x^2}}$.$x = 2 	an 	heta$ આદેશ લેતા,$dx = 2 \sec^2 	heta \ d	heta$ મળે.સંકલન $I = \int \frac{2 \sec^2 	heta \ d	heta}{(4 	an^2 	heta) \sqrt{4 + 4 	an^2 	heta}}$ થશે.અહીં $\sqrt{4(1 + 	an^2 	heta)} = 2 \sec 	heta$ હોવાથી,$I = \int \frac{2 \sec^2 	heta \ d	heta}{4 	an^2 	heta \cdot 2 \sec 	heta} = \frac{1}{4} \int \frac{\sec 	heta}{	an^2 	heta} \ d	heta$.$I = \frac{1}{4} \int \frac{1/\cos 	heta}{\sin^2 	heta / \cos^2 	heta} \ d	heta = \frac{1}{4} \int \frac{\cos 	heta}{\sin^2 	heta} \ d	heta$.$u = \sin 	heta$ લેતા,$du = \cos 	heta \ d	heta$ મળે. તેથી $I = \frac{1}{4} \int u^{-2} \ du = \frac{1}{4} (-u^{-1}) + C = -\frac{1}{4 \sin 	heta} + C$.$	an 	heta = \frac{x}{2}$ હોવાથી,$\sin 	heta = \frac{x}{\sqrt{x^2+4}}$ થાય.આમ,$I = -\frac{1}{4} \cdot \frac{\sqrt{x^2+4}}{x} + C = -\frac{\sqrt{4+x^2}}{4x} + C$.

$\int \frac{dx}{x^2 \sqrt{4+x^2}}$ ની કિંમત શોધો.

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