Correct option: D.$g - g ^{\prime}$ is increasing in $\left(0, \frac{\pi}{2}\right)$
d
$\int\left(\frac{x(\cos x-\sin x)}{e^{x}+1}+\frac{g(x)\left(e^{x}+1-x e^{x}\right)}{\left(e^{x}+1\right)^{2}}\right) d x=\frac{x g(x)}{e^{x}+1}+c$
On differentiating both sides w.r.t. $x$, we get
$\left(\frac{x(\cos x-\sin x)}{e^{x}+1}+\frac{g(x)\left(e^{x}+1-x e^{x}\right.}{\left(e^{x}+1\right)^{2}}\right)$
$=\frac{\left(e^{x}+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^{x} \cdot x \cdot g(x)}{\left(e^{x}+1\right)^{2}}$
$\left(e^{x}+1\right) x(\cos x-\sin x)+g(x)\left(e^{x}+1-x e^{x}\right)$
$=\left(e^{x}+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^{x} \cdot x \cdot g(x)$
$\Rightarrow g^{\prime}(x)=\cos x-\sin x$
$\Rightarrow g(x)=\sin x+\cos x+C$
$g ( x )$ is increasing in $(0, \pi / 4)$
$g ^{\prime \prime}( x )=-\sin x -\cos x <0$
$\Rightarrow g ^{\prime}( x )$ is decreasing function
let $h ( x )= g ( x )+ g ^{\prime}( x )=2 \cos x + C$
$\Rightarrow h ^{\prime}( x )= g ^{\prime}( x )+ g ^{\prime \prime}( x )=-2 \sin x <0$
$\Rightarrow h$ is decreasing
let $\phi( x )= g ( x )- g ^{\prime}( x )=2 \sin x + C$
$\Rightarrow \phi^{\prime}( x )= g ^{\prime}( x )- g { }^{\prime \prime}( x )=2 \cos x >0$
$\Rightarrow \phi$ is increasing