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Evaluate int(0)^(1)(e^(-x)dx)/(1+e^(x))...

Evaluate `int_(0)^(1)(e^(-x)dx)/(1+e^(x))`

Text Solution

Verified by Experts

The correct Answer is:
`log(1+e)-1/3-log2`
`I=int_(0)^(1)(e^(-x) dx)/(1+e^(x))=int_(0)^(1)(dx)/(e^(x)(1+e^(x)))`
Put `e^(x)=z`
`:.e^(x)dx=dz`
`implies dx=(dz)/(e^(x))=(dz)/z`
`impliesI=int_(1)^(e)(dz)/(z^(2)(1+z))`
`=int_(1)^(e)(1/(1+z)-(z-1)/(z^(2)))dz`
`=|log_(e)(1+z)-log_(e)z-1/z|_(1)^(e)`
`=(log_(e)(1+e)-log_(e) e-1/e)-(log_(e)2-log_(e)1-1)`
`=log(1+e)-1/e-log2`
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