No, its $\displaystyle x+\ln{(e^{-x}+1)$. It works out the same towards the end when you take log of both sides. I keep getting the same answer and including your answer. I wonder is the book wrong? See attached..
No, its $\displaystyle x+\ln{(e^{-x}+1)$. It works out the same towards the end when you take log of both sides. I keep getting the same answer and including your answer. I wonder is the book wrong? See attached..