A) 0
B) \[\pi /4\]
C) \[\pi /2\]
D) \[-\pi /4\]
Correct Answer: B
Solution :
\[I=\int_{-1}^{0}{\frac{dx}{{{x}^{2}}+2x+2}}\]\[=\int_{-1}^{0}{\frac{dx}{{{(x+1)}^{2}}+1}}\] \[=[{{\tan }^{-1}}(x+1)]_{-1}^{0}\]\[=[{{\tan }^{-1}}1-{{\tan }^{-1}}0]=\frac{\pi }{4}\].You need to login to perform this action.
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