Evaluate the following definite integrals : _0^1 1 1+x+x d x - Vidyadip

Question

Evaluate the following definite integrals : $\int_0^1 \frac{1}{\sqrt{1+x}+\sqrt{x}} d x$

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Answer

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

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