$\ln(xe^{-x}+e^{-x})=\ln(x+1)-x$?

Question

Actually it is small part of this question . I am unable to understand how $\ln(xe^{-x}+e^{-x})=\ln(x+1)-x$ ? Please give only hint.

Answer 1

we have $$\ln(x+1)-x=\ln(x+1)-\ln(e^x)=\ln\left(\frac{x+1}{e^x}\right)=\ln(xe^{-x}+e^{-x})$$