Let g: (0, ∞) → R be a differentiable function such that ∫ ( x( x -...

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

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

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

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

So $g\left(x\right)$ is increasing in \left(0, \dfrac{\pi}{4}\right)$</p><p>${g}^{'}\left(x\right) = \cos x - \sin x$</p><p>i.e.$g\left(x\right) - {g}^{'}\left(x\right) = 2\sin x$is an increasing function in \left(0, \dfrac{\pi}{2}\right)$.