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  4. If I=∫ e-x log (ex+1) d x, then I equals

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Q. If $I=\int e^{-x} \log \left(e^{x}+1\right) d x$, then $I$ equals

Integrals

Solution:

$I =-e^{-x} \log \left(e^{x}+1\right)+\int \frac{e^{-x} e^{x}}{e^{x}+1} d x$
$=-e^{-x} \log \left(e^{x}+1\right)+\int \frac{e^{-x}}{e^{-x}+1} d x$
$=-e^{-x} \log \left(e^{x}+1\right)-\log \left(e^{-x}+1\right)+C$
$=-e^{-x} \log \left(e^{x}+1\right)-\log \left(1+e^{x}\right)+x+ C$
$=-\left(e^{-x}+1\right) \log \left(e^{x}+1\right)+x+ C$