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

MCQ

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

A
$\frac{2 \sqrt{2}}{3}$
✓
$\frac{4 \sqrt{2}}{3}$
C
$\frac{8 \sqrt{2}}{3}$
D
$\frac{\sqrt{2}}{3}$

✓

Answer

Correct option: B.

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

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

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