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intcos(logx)dx=F(x)+c, where c is an arb...

`intcos(logx)dx=F(x)+c`, where c is an arbitrary constant. Here F(x)=

A

`x[cos(logx)+sin(logx)]`

B

`x[cos(logx)-sin(logx)]`

C

`x/2[cos(logx)+sin(logx)]`

D

`x/2[cos(logx)-sin(logx)]`

Text Solution

Verified by Experts

The correct Answer is:
C
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