_ ^ x 1 - x^2 1 + x^2 ;dx = - Vidyadip

MCQ

$\int_{}^{} {x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} } \;dx = $

✓
$\frac{1}{2}[{\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^4}} ] + c$
B
$\frac{1}{2}[{\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^2}} ] + c$
C
${\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^4}} + c$
D
${\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^2}} + c$

✓

Answer

Correct option: A.

$\frac{1}{2}[{\sin ^{ - 1}}{x^2} + \sqrt {1 - {x^4}} ] + c$

a
(a)$\int_{}^{} {x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} } dx = \int_{}^{} {\frac{{x.\,(1 - {x^2})}}{{\sqrt {1 - {x^4}} }}} dx$
$\{$ Multiplying ${N^r}$ and ${D^r}$ by ${(1 - {x^2})^{1/2}}\} $
$ = \int_{}^{} {\frac{x}{{\sqrt {1 - {x^4}} }}} \,dx - \int_{}^{} {\frac{{{x^3}}}{{\sqrt {1 - {x^4}} }}} dx$

$ = \frac{1}{2}[{\sin ^{ - 1}}({x^2}) + \sqrt {1 - {x^4}} ] + c$.
(By putting ${x^2} = t$ and $\sqrt {1 - {x^4}} = \sqrt t $respectively)

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