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If I=inte^(-x)log(e^x+1)dx ,t h e nIe q ...

If `I=inte^(-x)log(e^x+1)dx ,t h e nIe q u a l` `a+(e^(-x)+1)log(e^x+1)+C` `a+(e^x+1)log(e^x+1)+C` `a-(e^(-x)+1)log(e^x+1)+C` none of these

A

`x+(e^(-x)+1)log(e^(x)+1)+C`

B

`x+(e^(x)+1)log(e^(x)+1)+C`

C

`x-(e^(-x)+1)log(e^(x)+1)+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`I=-e^(-x)log(e^(x)+1)+int(e^(-x)e^(x))/(e^(x)+1)dx `
`=-e^(-x)log(e^(x)+1)+int(e^(-x))/(e^(-x)+1)dx`
`=-e^(-x)log(e^(x)+1)-log(e^(-x)+1)+C`
`=-e^(-x)log(e^(x)+1)-log(1+e^(x))+x+C`
`=-(e^(-x)+1)log(e^(x)+1)+x+C`
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