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

$$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c} $$, for all x > 0, where c is an arbitrary constant. Then :

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