$\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$ is equal to

Given: $\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$

The denominator has a difference of square roots. Multiply both numerator and denominator by the conjugate $\sqrt{1+x} + \sqrt{x}$:

$\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}} \times \frac{\sqrt{1+x} + \sqrt{x}}{\sqrt{1+x} + \sqrt{x}}$

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Using $(a-b)(a+b) = a^2 - b^2$:

Denominator:

$(\sqrt{1+x} - \sqrt{x})(\sqrt{1+x} + \sqrt{x})$

$= (\sqrt{1+x})^2 - (\sqrt{x})^2$

$= (1+x) - x$

$= 1$

Numerator: $\sqrt{1+x} + \sqrt{x}$

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The integral becomes:

$\int_0^1 (\sqrt{1+x} + \sqrt{x}) \, dx$

$= \int_0^1 \sqrt{1+x} \, dx + \int_0^1 \sqrt{x} \, dx$

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For $\int_0^1 \sqrt{1+x} \, dx$:

$\int_0^1 (1+x)^{1/2} \, dx$

$= \left[\frac{(1+x)^{3/2}}{3/2}\right]_0^1$

$= \left[\frac{2}{3}(1+x)^{3/2}\right]_0^1$

$= \frac{2}{3}(2)^{3/2} - \frac{2}{3}(1)^{3/2}$

$= \frac{2}{3} \times 2\sqrt{2} - \frac{2}{3}$

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

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For $\int_0^1 \sqrt{x} \, dx$:

$\int_0^1 x^{1/2} \, dx$

$= \left[\frac{x^{3/2}}{3/2}\right]_0^1$

$= \left[\frac{2}{3}x^{3/2}\right]_0^1$

$= \frac{2}{3}(1)^{3/2} - 0$

$= \frac{2}{3}$

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Adding both results:

$\left(\frac{4\sqrt{2}}{3} - \frac{2}{3}\right) + \frac{2}{3}$

$= \frac{4\sqrt{2}}{3} - \frac{2}{3} + \frac{2}{3}$

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

Therefore, $\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}} = \frac{4\sqrt{2}}{3}$