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01xexx+12dx

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a
e2
b
e−12
c
e2−1
d
e−32

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detailed solution

Correct option is C

G. I=∫01 (x+1)−1(x+1)2exdx=∫01 1x+1−1(x+1)2exdx=exx+101=e2−1


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