int frac{1}{sqrt{x}+x sqrt{x}} d x | vedclass

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Question

(B) Let $ I = \int \frac{1}{\sqrt{x} + x \sqrt{x}} dx $. Factor out $ \sqrt{x} $ from the denominator: $ I = \int \frac{1}{\sqrt{x}(1 + x)} dx $. Sinc

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $ I = \int \frac{1}{\sqrt{x} + x \sqrt{x}} dx $. Factor out $ \sqrt{x} $ from the denominator: $ I = \int \frac{1}{\sqrt{x}(1 + x)} dx $. Since $ x = (\sqrt{x})^2 $,we have $ I = \int \frac{1}{\sqrt{x}(1 + (\sqrt{x})^2)} dx $. Let $ t = \sqrt{x} $. Then $ dt = \frac{1}{2\sqrt{x}} dx $,which implies $ \frac{1}{\sqrt{x}} dx = 2 dt $. Substituting these into the integral: $ I = \int \frac{2}{1 + t^2} dt $. Integrating,we get $ I = 2 	an^{-1}(t) + C $. Substituting $ t = \sqrt{x} $ back,we get $ I = 2 	an^{-1}(\sqrt{x}) + C $.

$ \int \frac{1}{\sqrt{x}+x \sqrt{x}} d x $

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