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} $
(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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