_0^1 1 1+x+x d x - Vidyadip

Question

$\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$

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Answer

$
\begin{aligned}
& \text { Let } I =\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} \cdot d x \\
& =\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} \times \frac{\sqrt{1+x}-\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} \cdot d x \\
& =\int_0^1 \frac{\sqrt{1+x}-\sqrt{x}}{(\sqrt{1+x})^2-\left(\sqrt{x}^2\right) \cdot d x} \\
& =\int_0^1 \frac{\sqrt{1+x}-\sqrt{x}}{1+x-x} \cdot d x \\
& =\int_0^1\left[(1+x)^{\frac{1}{2}}-x^{\frac{1}{2}}\right] \cdot d x \\
& =\int_0^1(1+x)^{\frac{1}{2}} \cdot d x-\int_0^1 x^{\frac{1}{2}} \cdot d x \\
& =\left[\frac{(1+x)^{\frac{1}{2}}}{\frac{3}{2}}\right]_0^1-\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^1 \\
& =\frac{2}{3}\left[(2)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]-\frac{2}{3}\left[(1)^{\frac{3}{2}}-0\right] \\
& =\frac{2}{3}(2 \sqrt{2}-1)-\frac{2}{3}(1) \\
& =\frac{4 \sqrt{2}}{3}-\frac{2}{3}-\frac{2}{3} \\
& \therefore I =\frac{4}{2}(\sqrt{2}-1) \text {. } \\
\end{aligned}
$

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