_ ^ x ^ - 1 xdx = - Vidyadip

MCQ

$\int_{}^{} {x{{\tan }^{ - 1}}} xdx = $

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

✓

Answer

Correct option: A.

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

a
(a)$\int_{}^{} {x\,.\,{{\tan }^{ - 1}}x\,dx = \frac{{{x^2}}}{2}{{\tan }^{ - 1}}x - \frac{1}{2}\int_{}^{} {\frac{{{x^2} + 1 - 1}}{{1 + {x^2}}}\,dx} } $
$ = \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}x + \frac{1}{2}{\tan ^{ - 1}}x + c$
$ = \frac{1}{2}{\tan ^{ - 1}}x\,.\,({x^2} + 1) - \frac{1}{2}x + c$.

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