int frac{dx}{sqrt{x} + sqrt{x - 2}} = | vedclass

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(A) To solve the integral $I = \int \frac{dx}{\sqrt{x} + \sqrt{x - 2}}$,we rationalize the denominator by multiplying the numerator and denominator by

Vedclass · Accepted Answer

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

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(A) To solve the integral $I = \int \frac{dx}{\sqrt{x} + \sqrt{x - 2}}$,we rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{x} - \sqrt{x - 2})$.$I = \int \frac{\sqrt{x} - \sqrt{x - 2}}{(\sqrt{x} + \sqrt{x - 2})(\sqrt{x} - \sqrt{x - 2})} dx$$I = \int \frac{\sqrt{x} - \sqrt{x - 2}}{x - (x - 2)} dx$$I = \int \frac{\sqrt{x} - \sqrt{x - 2}}{2} dx$$I = \frac{1}{2} \int (x^{1/2} - (x - 2)^{1/2}) dx$Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$,we get:$I = \frac{1}{2} \left[ \frac{x^{3/2}}{3/2} - \frac{(x - 2)^{3/2}}{3/2} \right] + c$$I = \frac{1}{2} \cdot \frac{2}{3} [x^{3/2} - (x - 2)^{3/2}] + c$$I = \frac{1}{3} [x^{3/2} - (x - 2)^{3/2}] + c$

$\int \frac{dx}{\sqrt{x} + \sqrt{x - 2}} = $

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