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i=int(e^(x))/((x+2)){1+(x+2)log(x+2)}dx=...

`i=int(e^(x))/((x+2)){1+(x+2)log(x+2)}dx=`

A

`e^(x).log(x+2)+c`

B

`e^(x)//(x+2)+c`

C

`e^(x)(x+2)+c`

D

`e^(x)(x-2)+c`

Text Solution

Verified by Experts

The correct Answer is:
A

`1= int e^(x).[(1)/(underset(g'(x))(ubrace(x+2)))+underset(g(x))(ubrace(log(x+2))]]dx`
`rArr 1= e^(x). g(x) =e^(x). log (x+2) c`
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