_ ^ x ^ - 1 x (1 + x^2 ) ^ 3/2 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{x{{\tan }^{ - 1}}x}}{{{{(1 + {x^2})}^{3/2}}}}\;dx = } $

A
$\frac{{x + {{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }} + c$
✓
$\frac{{x - {{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }} + c$
C
$\frac{{{{\tan }^{ - 1}}x - x}}{{\sqrt {1 + {x^2}} }} + c$
D
None of these

✓

Answer

Correct option: B.

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

b
(b) Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$ then
$\int_{}^{} {\frac{{x{{\tan }^{ - 1}}x}}{{{{(1 + {x^2})}^{32}}}}\,dx} = \int_{}^{} {\frac{{\theta \tan \theta {{\sec }^2}\theta \,d\theta }}{{{{(1 + {{\tan }^2}\theta )}^{32}}}}} $
$ = \int_{}^{} {\theta \sin \theta \,d\theta } = - \theta \cos \theta + \sin \theta + c$
$ = \frac{x}{{\sqrt {{x^2} + 1} }} - {\tan ^{ - 1}}x\frac{1}{{\sqrt {{x^2} + 1} }} = \frac{{x - {{\tan }^{ - 1}}x}}{{\sqrt {1 + {x^2}} }} + c$.

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