MCQ
$\int_{\, - \,1}^{\,0} {\frac{{dx}}{{{x^2} + 2x + 2}} = } $
- A$0$
- ✓$\pi /4$
- C$\pi /2$
- D$ - \pi /4$
✓
Answer
Correct option: B.
$\pi /4$
b
(b) $I = \int_{ - 1}^0 {\frac{{dx}}{{{x^2} + 2x + 2}}} $$ = \int_{ - 1}^0 {\frac{{dx}}{{{{(x + 1)}^2} + 1}}} $
(b) $I = \int_{ - 1}^0 {\frac{{dx}}{{{x^2} + 2x + 2}}} $$ = \int_{ - 1}^0 {\frac{{dx}}{{{{(x + 1)}^2} + 1}}} $
$ = [{\tan ^{ - 1}}(x + 1)]_{ - 1}^0$
$ = [{\tan ^{ - 1}}1 - {\tan ^{ - 1}}0] = \frac{\pi }{4}$.
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