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.
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Hint: $\ln(ab)=\ln(a)+\ln(b)$ and $\ln(e^{-x})=-x$. – lulu Mar 14 '17 at 12:44
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try looking at the common factor and then use the properties of the logarithm – user96233 Mar 14 '17 at 12:45
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2$\ln(xe^{-x}+e^{-x})=\ln\bigl((x+1) e^{-x}\bigr)=\ln(x+1)+\ln (e^{-x})$. – Bernard Mar 14 '17 at 12:45
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we have $$\ln(x+1)-x=\ln(x+1)-\ln(e^x)=\ln\left(\frac{x+1}{e^x}\right)=\ln(xe^{-x}+e^{-x})$$
Dr. Sonnhard Graubner
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you are right, it was a typo,sorry thank you for the hint – Dr. Sonnhard Graubner Mar 14 '17 at 13:12