_ ,0 ^ ,1 , ^ - 1 ( 1 x^2 - x + 1 ) ,dx is - Vidyadip

MCQ

$\int_{\,0}^{\,1} {\,{{\tan }^{ - 1}}\left( {\frac{1}{{{x^2} - x + 1}}} \right)\,dx} $ is

A
$ln\ 2$
B
$ - \ln 2$
C
$\frac{\pi }{2} + \ln 2$
✓
$\frac{\pi }{2} - \ln 2$

✓

Answer

Correct option: D.

$\frac{\pi }{2} - \ln 2$

d
(d) $\int_0^1 {{{\tan }^{ - 1}}\left( {\frac{1}{{{x^2} - x + 1}}} \right)\,dx} $

$= \int_0^1 {{{\tan }^{ - 1}}x\,dx - } \int_0^1 {{{\tan }^{ - 1}}(x - 1)} \,dx$

$ = 2\int_{\,0}^{\,1} {{{\tan }^{ - 1}}x\,dx} $

$= 2\,[{\tan ^{ - 1}}x - \frac{1}{2}\log (1 + {x^2})]_0^1 $

$= \frac{\pi }{2} - \log 2.$

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