1. Simplify the functions

Given $f(x)=\ln(x^2+1)-e^{-x}+1$ and $g(x)=\frac{1-2e^{2x}}{e^x}=e^{-x}-2e^x.$

We need to solve $f\left(g\left(\frac{(\alpha-1)^2}{3}\right)\right)>f\left(g\left(\alpha-\frac53\right)\right).$

2. Study monotonicity of $f$

Differentiate: $f'(x)=\frac{2x}{x^2+1}+e^{-x}.$

Now check whether $f'(x)>0$ for all real $x$.

Let $h(x)=\frac{2x}{x^2+1}+e^{-x}.$

• If $x\ge 0$, then both terms are nonnegative and $e^{-x}>0$, so $h(x)>0$.
• If $x<0$, write $x=-t$ where $t>0$. Then $h(-t)=-\frac{2t}{t^2+1}+e^t.$ Since $e^t>1\ge \frac{2t}{t^2+1}$ (because $2t\le t^2+1 \iff (t-1)^2\ge 0$), we get $h(-t)>0$.

Hence, $f'(x)>0\quad \forall x\in\mathbb R.$ So $f$ is strictly increasing on $\mathbb R$.

Therefore, $f(u)>f(v) \iff u>v.$

So the given inequality becomes $g\left(\frac{(\alpha-1)^2}{3}\right)>g\left(\alpha-\frac53\right).$

3. Study monotonicity of $g$

Recall $g(x)=e^{-x}-2e^x.$ Differentiate: $g'(x)=-e^{-x}-2e^x<0 \quad \forall x\in\mathbb R.$ Thus $g$ is strictly decreasing on $\mathbb R$.

Therefore, $g(u)>g(v) \iff u<v.$

So we need $\frac{(\alpha-1)^2}{3}<\alpha-\frac53.$

4. Solve the quadratic inequality

Multiply by $3$: $(\alpha-1)^2<3\alpha-5.$ Expand: $\alpha^2-2\alpha+1<3\alpha-5.$ $\alpha^2-5\alpha+6<0.$ Factor: $(\alpha-2)(\alpha-3)<0.$

Hence, $2<\alpha<3.$

5. Match with options

The correct range is $\boxed{(2,3)}$ which is Option A.

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.