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aalogx+x2+1dx=

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a
2 log a2+1
b
2 log a2+1−a
c
0
d
2 log a+a2+1

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detailed solution

Correct option is C

I=∫−aa log⁡x+x2+1dx=∫−aa log⁡−x+x2+1dx∫−aa f(x)dx=∫−aa f(−x)dx∫−aa log⁡−x2+x2+1x+x2+1dx=−∫−aa log⁡x+x2+1dx=−I⇒I=0


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