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Q.

∫01xexx+12dx

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a

e2

b

e−12

c

e2−1

d

e−32

answer is C.

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Detailed Solution

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