_ , - ,1 ^ ,0 dx x^2 + 2x + 2 = - Vidyadip

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}}} $

$ = [{\tan ^{ - 1}}(x + 1)]_{ - 1}^0$

$ = [{\tan ^{ - 1}}1 - {\tan ^{ - 1}}0] = \frac{\pi }{4}$.

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