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

MCQ

$\int_{}^{} {{{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}dx = } $

A
$x{\tan ^{ - 1}}x + c$
B
$x{\tan ^{ - 1}}x - \log (1 + {x^2}) + c$
C
$2x{\tan ^{ - 1}}x + \log (1 + {x^2}) + c$
✓
$2x{\tan ^{ - 1}}x - \log (1 + {x^2}) + c$

✓

Answer

Correct option: D.

$2x{\tan ^{ - 1}}x - \log (1 + {x^2}) + c$

d
(d) Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$ then
$\int_{}^{} {{{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\,dx} = \int_{}^{} {{{\tan }^{ - 1}}\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \,\,\,{\sec ^2}\theta \,d\theta $
$ = \int_{}^{} {{{\tan }^{ - 1}}(\tan 2\theta ){{\sec }^2}\theta \,d\theta } = \int_{}^{} {2\theta {{\sec }^2}\theta \,d\theta } $
$ = 2\left[ {\theta \tan \theta - \int_{}^{} {\tan \theta \,d\theta } } \right]$$ = 2x{\tan ^{ - 1}}x - \log ({x^2} + 1) + c.$

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