_ ^ 1 - x 1 + x ;dx = - Vidyadip

MCQ

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

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

✓

Answer

Correct option: D.

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

d
(d)$\int_{}^{} {\sqrt {\frac{{1 - x}}{{1 + x}}} \,dx} = \int_{}^{} {\frac{{1 - x}}{{\sqrt {1 - {x^2}} }}} \,dx = \int_{}^{} {\frac{1}{{\sqrt {1 - {x^2}} }}} \,dx - \int_{}^{} {\frac{{x\,dx}}{{\sqrt {1 - {x^2}} }}} $
Now proceed yourself.

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