Let $g:(0,\infty )\to \mathrm{R}$ be a differentiable function such that

$\int \left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g\left(x\right)\left(e^x+1-x{e}^x\right)}{{\left(e^x+1\right)}^2}\right)dx=\frac{xg\left(x\right)}{e^x+1}+c$

for all $\mathrm{x}>0$, where $\mathrm{c}$ is an arbitrary constant. Then.

$\int \left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g\left(x\right)\left(e^x+1-x{e}^x\right)}{{\left(e^x+1\right)}^2}\right)dx=\frac{xg\left(x\right)}{e^x+1}+c$

On differentiating both sides w.r.t. x, we get

$$\begin{gathered} \left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g\left(x\right)\left(e^x+1-x{e}^x\right.}{{\left(e^x+1\right)}^2}\right) \\ =\frac{\left(e^x+1\right)\left(g\left(x\right)+x{g}^{\text{'}}\left(x\right)\right)-e^x·x·g\left(x\right)}{{\left(e^x+1\right)}^2} \\ \Rightarrow \left(e^x+1\right)x(\cos x-\sin x)+g\left(x\right)\left(e^x+1-x{e}^x\right) \\ =\left(e^x+1\right)\left(g\left(x\right)+x{g}^{\text{'}}\left(x\right)\right)-e^x·x·g\left(x\right) \\ \Rightarrow g^{\text{'}}\left(x\right)=\cos x-\sin x \\ \Rightarrow g\left(x\right)=\sin x+\cos x+C \\ \therefore \mathrm{g}\left(\mathrm{x}\right)\text{ is increasing in }(0,\pi /4) \\ \Rightarrow g^{\text{'}\text{'}}\left(x\right)=-\sin x-\cos x<0 \\ \Rightarrow {\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)\text{ is decreasing function } \\ \text{ let }h\left(x\right)=g\left(x\right)+g^{\text{'}}\left(x\right)=2\cos x+C \\ \Rightarrow h^{\text{'}}\left(x\right)=g^{\text{'}}\left(x\right)+g^{\text{'}\text{'}}\left(x\right)=-2\sin x<0 \\ \Rightarrow \mathrm{h}\text{ is decreasing } \\ \text{ let }\varphi \left(x\right)=g\left(x\right)-g^{\text{'}}\left(x\right)=2\sin x+C \\ \Rightarrow {\varphi }^{\text{'}}\left(x\right)=g^{\text{'}}\left(x\right)-g^{\text{'}\text{'}}\left(x\right)=2\cos x>0 \\ \Rightarrow \varphi \text{ is increasing } \end{gathered}$$

Hence option D is correct.