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Question

Question: \(\int _ { - 1 } ^ { 0 } \frac { d x } { x ^ { 2 } + 2 x + 2 } =\)...

∫−10dxx2+2x+2=\int _ { - 1 } ^ { 0 } \frac { d x } { x ^ { 2 } + 2 x + 2 } =

A

0

B

π/4\pi / 4

C

π/2\pi / 2

D

−π/4- \pi / 4

Answer

π/4\pi / 4

Explanation

Solution

I=∫−10dxx2+2x+2I = \int _ { - 1 } ^ { 0 } \frac { d x } { x ^ { 2 } + 2 x + 2 } =∫−10dx(x+1)2+1= \int _ { - 1 } ^ { 0 } \frac { d x } { ( x + 1 ) ^ { 2 } + 1 }

=[tan⁡−1(x+1)]−10= \left[ \tan ^ { - 1 } ( x + 1 ) \right] _ { - 1 } ^ { 0 } =[tan⁡−11−tan⁡−10]=π4= \left[ \tan ^ { - 1 } 1 - \tan ^ { - 1 } 0 \right] = \frac { \pi } { 4 }.