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11x3+|x|+1x2+2|x|+1dx is equal to

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a
log 2
b
2 log 2
c
12 log 2
d
4 log 2

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detailed solution

Correct option is B

Since,I=∫−11 x3+|x|+1x2+2|x|+1dxI=∫−11 x3x2+2|x|+1+∫−11 |x|+1x2+2|x|+1dx=0+2∫01 |x|+1(|x|+1)2dx    [odd function + even function]=2∫01 x+1(x+1)2dx=2∫01 1x+1dx=2[log∣x+1∥]01=2log⁡2


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