_ ^ dx 2 x (1 + x) = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{2\sqrt x (1 + x)}} = } $

A
$\frac{1}{2}{\tan ^{ - 1}}(\sqrt x ) + c$
✓
${\tan ^{ - 1}}(\sqrt x ) + c$
C
$2{\tan ^{ - 1}}(\sqrt x ) + c$
D
None of these

✓

Answer

Correct option: B.

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

b
(b) $I = \int {\frac{{dx}}{{2\sqrt x (1 + x)}}} $

Put $\sqrt x \, = t$==> $\frac{1}{{2\sqrt x }}dx = dt$

$\therefore I = \int {\frac{{dt}}{{1 + {t^2}}}} = {\tan ^{ - 1}}t + c$$ = {\tan ^{ - 1}}(\sqrt x ) + c$.

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