int frac{dx}{sqrt{1 + x} + sqrt{x}} = | vedclass

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Question

(D) To solve the integral $\int \frac{dx}{\sqrt{1 + x} + \sqrt{x}}$,we rationalize the denominator by multiplying the numerator and denominator by $(\

Vedclass · Accepted Answer

$\frac{2}{3}(1 + x)^{3/2} - \frac{2}{3}x^{3/2} + c$

Vedclass · Answer

(D) To solve the integral $\int \frac{dx}{\sqrt{1 + x} + \sqrt{x}}$,we rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{1 + x} - \sqrt{x})$.$\int \frac{1}{\sqrt{1 + x} + \sqrt{x}} 	imes \frac{\sqrt{1 + x} - \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}} dx$$= \int \frac{\sqrt{1 + x} - \sqrt{x}}{(1 + x) - x} dx$$= \int (\sqrt{1 + x} - \sqrt{x}) dx$$= \int (1 + x)^{1/2} dx - \int x^{1/2} dx$$= \frac{(1 + x)^{3/2}}{3/2} - \frac{x^{3/2}}{3/2} + c$$= \frac{2}{3}(1 + x)^{3/2} - \frac{2}{3}x^{3/2} + c$.

$\int \frac{dx}{\sqrt{1 + x} + \sqrt{x}} = $

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