int[sin(logx)+cos(logx)]dx...

`int[sin(logx)+cos(logx)]dx`

A

`x log (logx)+C`

B

`sin(logx)+C`

C

`cos(logx)+C`

D

`x sin (logx)+C`

Text Solution

Verified by Experts

The correct Answer is:

D

`int[sin(logx)+cos(logx)]dx`
`=intsin(logx)dx+intcos(logx)dx`
`=x sin (logx)-int(xcos (logx))/(x)dx+intcos(logx)dx+C`
`=x sin (logx)+C`

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