Evaluate the definite integral: int_0^1 frac{ | vedclass

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(B) Let $I = \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+x} + \sqr

Vedclass · Accepted Answer

$\frac{4\sqrt{2}}{3}$

Vedclass · Answer

(B) Let $I = \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+x} + \sqrt{x})$:$I = \int_0^1 \frac{\sqrt{1+x} + \sqrt{x}}{(1+x) - x} dx = \int_0^1 (\sqrt{1+x} + \sqrt{x}) dx$.Integrating term by term:$I = \left[ \frac{2}{3}(1+x)^{3/2} + \frac{2}{3}x^{3/2} \right]_0^1$.Evaluating at the limits:$I = \left( \frac{2}{3}(2)^{3/2} + \frac{2}{3}(1)^{3/2} \right) - \left( \frac{2}{3}(1)^{3/2} + 0 \right)$.$I = \frac{2}{3}(2\sqrt{2}) + \frac{2}{3} - \frac{2}{3} = \frac{4\sqrt{2}}{3}$.

Evaluate the definite integral: $\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$

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