$\int e^{-x}\log \left(e^x+1\right)dx=x-Q\left(x\right)\log h\left(x\right)-\log \left(h\right(x\left)\right)+C\text{ than }$

$$\begin{gathered} I=\int \frac{1}{{\left(e^x\right)}^2}\log \left(e^x+1\right)\left(e^{x}dx\right) \\ \text{put }{e}^x=t \\ I=\int \log \left(t+1\right).\frac{1}{t^2}dt \\ =-\frac{\log \left(t+1\right)}{t}+\int \frac{1}{t+1}\left(\frac{-1}{t}\right)dt \\ =-\frac{\log \left(t+1\right)}{t}+\int \left[\frac{1}{t}-\frac{1}{t+1}\right]dt \\ =-\frac{\log \left(t+1\right)}{t}+\log t-\log \left(t+1\right)+c \\ =-\frac{\log \left(e^x+1\right)}{e^x}+x-\log \left(e^x+1\right)+c \end{gathered}$$