_ ^ x 1 + x dx = - Vidyadip

MCQ

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

A
$\sqrt x - {\tan ^{ - 1}}\sqrt x + c$
✓
$2(\sqrt x - {\tan ^{ - 1}}\sqrt x ) + c$
C
$2(\sqrt x + {\tan ^{ - 1}}x) + c$
D
$\sqrt {1 + x} + c$

✓

Answer

Correct option: B.

$2(\sqrt x - {\tan ^{ - 1}}\sqrt x ) + c$

b
(b)$\int_{}^{} {\frac{{\sqrt x }}{{1 + x}}\,dx = \int_{}^{} {\frac{{\sqrt x \sqrt x }}{{\sqrt x (1 + x)}}\,dx} } $
$ = \int_{}^{} {\frac{{x + 1}}{{\sqrt x (x + 1)}}\,dx - \int_{}^{} {\frac{1}{{\sqrt x (x + 1)}}\,dx} } $
$ = \int_{}^{} {\frac{1}{{\sqrt x }}\,dx - \int_{}^{} {\frac{1}{{\sqrt x (x + 1)}}\,dx} } $
$ = 2{x^{1/2}} - 2{\tan ^{ - 1}}\sqrt x + c = 2(\sqrt x - {\tan ^{ - 1}}\sqrt x ) + c.$

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