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inte^(-logx)dx is equal to...

`inte^(-logx)dx` is equal to

A

`e^(-logx)+C`

B

`-xe^(-logx)+C`

C

`e^(logx)+C`

D

`log|x|+C`

Text Solution

Verified by Experts

The correct Answer is:
D
Let `l=inte^(-logx)dx=int(1)/(x)dx=log|x|+C`
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