Evaluate the definite integral _ 0 ^ 1 d x 1+x-x - Vidyadip

Question

Evaluate the definite integral $\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$

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Answer

Integral to be evaluated: $\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$
Let $I=\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$
= $\int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} \times \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}} d x$
= $\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} d x$
= $\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1} d x$
= $\int_{0}^{1} \sqrt{1+x} d x+\int_{0}^{1} \sqrt{x} d x$
= $\int_{0}^{1}\left((1+x)^{\frac{1}{2}}\right) d x+\int_{0}^{1}(x)^{\frac{1}{2}} d x$
$\Rightarrow \mathrm{I}=\left[\frac{(1+\mathrm{x})^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}+\left[\frac{(\mathrm{x})^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}$
= $\frac{2}{3} \cdot\left[(1+1)^{\frac{3}{2}}-(1+0)^{\frac{3}{2}}\right]+\frac{2}{3} \cdot\left[(1)^{\frac{3}{2}}\right]$
= $\frac{2}{3} \cdot\left[(2)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]+\frac{2}{3} \cdot\left[(1)^{\frac{3}{2}}\right]$
= $\frac{2}{3} \cdot\left[(2)^{\frac{3}{2}}-1\right]+\frac{2}{3} \cdot[1]$
= $\frac{2}{3} \cdot[2 \sqrt{2}]$
$\Rightarrow$ $I=\frac{4 \sqrt{2}}{3}$

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