_ ^ dx 1 + x + x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{\sqrt {1 + x} + \sqrt x }} = } $

A
$\frac{2}{3}{(1 + x)^{2/3}} - \frac{2}{3}{x^{2/3}} + c$
B
$\frac{3}{2}{(1 + x)^{2/3}} + \frac{3}{2}{x^{2/3}} + c$
C
$\frac{3}{2}{(1 + x)^{3/2}} + \frac{3}{2}{x^{3/2}} + c$
✓
$\frac{2}{3}{(1 + x)^{3/2}} - \frac{2}{3}{x^{3/2}} + c$

✓

Answer

Correct option: D.

$\frac{2}{3}{(1 + x)^{3/2}} - \frac{2}{3}{x^{3/2}} + c$

d
(d)$\int_{}^{} {\frac{{dx}}{{\sqrt {1 + x} + \sqrt x }} = \int_{}^{} {\left[ {\frac{{(x + 1) - x}}{{\sqrt {1 + x} + \sqrt x }}} \right]\,dx} } $
$\int_{}^{} {(\sqrt {x + 1} - \sqrt x )\,dx} = \frac{{{{(x + 1)}^{3/2}}}}{{3/2}} - \frac{{{x^{3/2}}}}{{3/2}} + c$
$ = \frac{2}{3}[{(x + 1)^{3/2}} - {x^{3/2}}] + c = \frac{2}{3}{(x + 1)^{3/2}} - \frac{2}{3}{x^{3/2}} + c.$

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