Let $$g : (0, \infty) \to R$$ be a differentiable function such that $$\int \frac{x\cos x - \sin x}{e^x + 1} + \frac{g(x)e^x + 1 - xe^x}{(e^x + 1)^2} dx = \frac{xg(x)}{e^x + 1} + C$$, for all $$x > 0$$, where $$C$$ is an arbitrary constant. Then

Given,

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

Differentiating both sides with respect to $$x,$$

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

Using quotient rule on the RHS,

$$\frac{d}{dx}\left(\frac{xg(x)}{e^x+1}\right)=\frac{(e^x+1)(g(x)+xg'(x))-xe^xg(x)}{(e^x+1)^2}$$

Therefore,

$$\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)(e^x+1-xe^x)}{(e^x+1)^2}=\frac{(e^x+1)(g(x)+xg'(x))-xe^xg(x)}{(e^x+1)^2}$$

Multiplying throughout by $$(e^x+1)^2,$$

$$x(\cos x-\sin x)(e^x+1)+g(x)(e^x+1-xe^x)=(e^x+1)(g(x)+xg'(x))-xe^xg(x)$$

Now expand the RHS,

$$=(e^x+1)g(x)+x(e^x+1)g'(x)-xe^xg(x)$$

Observe that the terms containing $$g(x)$$ cancel from both sides. Hence,

$$x(\cos x-\sin x)(e^x+1)=x(e^x+1)g'(x)$$

Since $$x>0$$ and $$e^x+1>0,$$ dividing throughout gives

$$g'(x)=\cos x-\sin x$$

Integrating,

$$g(x)=\sin x+\cos x+C$$

Now consider

$$\phi(x)=g(x)-g'(x)$$

Substituting the values of $$g(x)$$ and $$g'(x),$$

$$\phi(x)=(\sin x+\cos x+C)-(\cos x-\sin x)$$

$$\phi(x)=2\sin x+C$$

Differentiating,

$$\phi'(x)=2\cos x$$

For $$0<x<\frac{\pi}{2},$$

$$\cos x>0$$

Hence,

$$\phi'(x)>0$$

Therefore, $$\phi(x)=g(x)-g'(x)$$ is increasing in $$\left(0,\frac{\pi}{2}\right)$$.