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xexdx is equal to 

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a
ex+C
b
2xex+C
c
(2x−4x+4)ex+C
d
None of these

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detailed solution

Correct option is C

Let,=∫xexdxPut x=t⇒12xdx=dt∴ I=2∫t2etdt=2t2et−2∫tetdt=2t2et−2tet−et+C=2t2et−2tet+2et+C=(2x−4x+4)ex+C


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