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

MCQ

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

A
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - {x^2}} }}{{x\sqrt 2 }}} \right] + c$
✓
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left[ {\frac{{x\sqrt 2 }}{{\sqrt {1 - {x^2}} }}} \right] + c$
C
$\sqrt 2 {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - {x^2}} }}{{x\sqrt 2 }}} \right] + c$
D
$ - \sqrt 2 {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - {x^2}} }}{{x\sqrt 2 }}} \right] + c$

✓

Answer

Correct option: B.

$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left[ {\frac{{x\sqrt 2 }}{{\sqrt {1 - {x^2}} }}} \right] + c$

b
(b) Put $x = \sin \theta \Rightarrow dx = \cos \theta \,d\theta ,$ then
$\int_{}^{} {\frac{{dx}}{{(1 + {x^2})\sqrt {1 - {x^2}} }} = \int_{}^{} {\frac{1}{{1 + {{\sin }^2}\theta }}\,d\theta } } = \int_{}^{} {\frac{{{{\sec }^2}\theta }}{{1 + 2{{\tan }^2}\theta }}\,d\theta } $
Again put $t = \tan \theta \Rightarrow dt = {\sec ^2}\theta \,d\theta ,$ then it reduces to
$\int_{}^{} {\frac{1}{{1 + 2{t^2}}}\,dt} = \frac{1}{2}\int_{}^{} {\frac{1}{{{t^2} + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}\,dt} $
$ = \frac{1}{2}\left( {\frac{1}{{(1/\sqrt 2 )}}} \right){\tan ^{ - 1}}\left( {\frac{t}{{(1/\sqrt 2 )}}} \right) + c$
$ = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan \theta ) + c = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{x\sqrt 2 }}{{\sqrt {1 - {x^2}} }}} \right) + c.$
Aliter : Put first $x = \frac{1}{t}$ and then ${t^2} - 1 = {z^2}.$

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