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(D) Given the integral equation: $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} \right) dx = \frac{x g(x)

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$g - g'$ is increasing in $(0, \pi/2)$

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(D) Given the integral equation: $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} \right) dx = \frac{x g(x)}{e^x + 1} + c$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus: $\frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} = \frac{d}{dx} \left( \frac{x g(x)}{e^x + 1} \right)$. Applying the quotient rule on the right side: $\frac{d}{dx} \left( \frac{x g(x)}{e^x + 1} \right) = \frac{(e^x + 1)(g(x) + x g'(x)) - x g(x) e^x}{(e^x + 1)^2} = \frac{(e^x + 1)g(x) + x g'(x)(e^x + 1) - x g(x) e^x}{(e^x + 1)^2} = \frac{g(x)(e^x + 1 - x e^x) + x g'(x)(e^x + 1)}{(e^x + 1)^2}$. Equating the terms: $\frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} = \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} + \frac{x g'(x)(e^x + 1)}{(e^x + 1)^2}$. Subtracting the common term $\frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2}$ from both sides: $\frac{x(\cos x - \sin x)}{e^x + 1} = \frac{x g'(x)}{e^x + 1}$. Since $x > 0$,we have $g'(x) = \cos x - \sin x$. Now,let's analyze the options: $1$. $g'(x) = \cos x - \sin x$. For $x \in (0, \pi/4)$,$\cos x > \sin x$,so $g'(x) > 0$,meaning $g$ is increasing. $2$. $g''(x) = -\sin x - \cos x$. For $x \in (0, \pi/4)$,$g''(x) $3$. Let $h(x) = g(x) + g'(x)$. Then $h'(x) = g'(x) + g''(x) = (\cos x - \sin x) + (-\sin x - \cos x) = -2 \sin x$. Since $-2 \sin x $4$. Let $\phi(x) = g(x) - g'(x)$. Then $\phi'(x) = g'(x) - g''(x) = (\cos x - \sin x) - (-\sin x - \cos x) = 2 \cos x$. Since $2 \cos x > 0$ for $x \in (0, \pi/2)$,$\phi$ is increasing. Thus,option $D$ is correct.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

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

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