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Question

exdx=f(x)+C; x>0. Then f(x) is

A
2(x1)ex
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B
1x)ex
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C
2(1x)ex
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D
(x1)ex
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Solution

The correct option is A 2(x1)ex
exdx=f(x)+C; x>0

Let x=t2, then dx=2t dt
exdx=et2t dt

Using integration by parts,
2t et dt=2t et2et dt
=2t et2et+C=2et(t1)+C
=2ex(x1)+C

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