Evaluate the definite integral in Exercise: ^ 1 _ 0 dx 1+x - x - Vidyadip

Question

Evaluate the definite integral in Exercise:
$\int^{1}_{0}\frac{\text{dx}}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}$

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Answer

$\text{Let I}=\int^{1}\limits_{0}\frac{\text{dx}}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}$
$\text{I}=\int^{1}\limits_{0}\frac{1}{\Big(\sqrt{1+\text{x}}-\sqrt{\text{x}}\Big)}\times\frac{\Big(\sqrt{1+\text{x}}+\sqrt{\text{x}}\Big)}{\Big(\sqrt{1+\text{x}}+\sqrt{\text{x}}\Big)}\text{dx}$
$=\int^{1}_{0}\frac{\sqrt{1+\text{x}}+\sqrt{\text{x}}}{1+\text{x}-\text{x}}\text{dx}$
$=\int^{1}_{0}{\sqrt{1+\text{x}}\text{dx}+\int^{1}\limits_{0}\sqrt{\text{x}}}\ \text{dx}$
$=\bigg[\frac{2}{3}\big(1+\text{x}\big)^{\frac{3}{2}}\bigg]^{1}_{0}+\bigg[\frac{2}{3}(\text{x)}^{\frac{3}{2}}\bigg]^{1}_{0}$
$=\frac{2}{3}\bigg[\big(2)^{\frac{3}{2}}-1\bigg]+\frac{2}{3}[1]$
$=\frac{2}{3}(2)^{\frac{3}{2}}$
$=\frac{2.2\sqrt{2}}{3}$
$=\frac{4\sqrt{2}}{3}$

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