1 + x 1 - x , ,dx = - Vidyadip

MCQ

$\int {\sqrt {\frac{{1 + x}}{{1 - x}}} \,\,dx = } $

A
$ - {\sin ^{ - 1}}x - \sqrt {1 - {x^2}} \, + c$
B
${\sin ^{ - 1}}x + \sqrt {1 - {x^2}} \, + c$
✓
${\sin ^{ - 1}}x - \sqrt {1 - {x^2}} \, + c$
D
$ - {\sin ^{ - 1}}x - \sqrt {{x^2} - 1} \, + c$

✓

Answer

Correct option: C.

${\sin ^{ - 1}}x - \sqrt {1 - {x^2}} \, + c$

c
(c)$I = \int {\sqrt {\frac{{1 + x}}{{1 - x}}} dx} $$ = \int {\frac{{1 + x}}{{\sqrt {1 - {x^2}} }}dx} $
$ = \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }} + \int {\frac{x}{{\sqrt {1 - {x^2}} }}dx} } $$ = {\sin ^{ - 1}}x - \sqrt {1 - {x^2}} + c$.

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