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1+xx1ex+x1dx is equal to 

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a
(x+1)ex+x−1+C
b
(x−1)ex+x−1+C
c
−xex+x−1+C
d
xex+x−1+C

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detailed solution

Correct option is D

∫1+x−x−1ex+x−1dx =∫xex+x−11−1x2+ex+x−1dx =xex+x−1−∫ex+x−1dx+∫ex+x−1dx =xex+x−1+C


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