Let g : 0 , ∞ → R be a differentiable function such that ∫ x cos x - sin x e x + 1 + g x e…

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

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

By comparison, we get, $g\left(x\right)=\sin x+\cos x$
$\Rightarrow g\left(x\right)=\sqrt{2}\left[\sin \left(x+\frac{\pi }{4}\right)\right]$

Since $x\in \left(0,\frac{\pi }{4}\right)$ so, $x+\frac{\pi }{4}\in \left(\frac{\pi }{4},\frac{\pi }{2}\right)$

So $g\left(x\right)$ is increasing in $\left(0,\frac{\pi }{4}\right)$

$g^{\text{'}}\left(x\right)=\cos x-\sin x$

i.e. $g\left(x\right)-g^{\text{'}}\left(x\right)=2\sin x$ is an increasing function in $\left(0,\frac{\mathrm{\pi }}{2}\right)$.