Let g:(0,)R be a differentiable function such that [[(x( x- x))/(e^x+1)+(g(x)(e^x+1-xe^x))…

Question

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

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

for all $x>0$, where $c$ is an arbitrary constant. Then:

Accepted Answer

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

On differentiating both sides w.r.t. $\mathrm{x}$, we get

$$
\begin{aligned}
&\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right.}{\left(e^x+1\right)^2}\right) \\
&=\frac{\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x)}{\left(e^x+1\right)^2} \\
&\left(e^x+1\right) x(\cos x-\sin x)+g(x)\left(e^x+1-x e^x\right) \\
&=\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x) \\
&\Rightarrow g g^{\prime}(x)=\cos x-\sin x \\
&\Rightarrow g(x)=\sin x+\cos x+C
\end{aligned}
$$

$\mathrm{g}(\mathrm{x})$ is increasing in $(0, \pi / 4)$

$g "(x)=-\sin x-\cos x0
$$

$

Solutions · 1