Let $f(x)=[1+x]+\frac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}.$ We need the left hand limit at $x=0$ and it is given to be equal to $\alpha-\frac43.$ We must find the integral value of $\alpha$.

1. Behaviour of $[x],\{x\},[1+x]$ as $x\to 0^-$

For $x\to 0^-$, we have $-1<x<0$. Therefore, $[x]=-1,\qquad \{x\}=x-[x]=x+1.$ Also, $1+x\in(0,1)\implies [1+x]=0.$

So near $x=0^-$, $2[x]+\{x\}=2(-1)+(x+1)=x-1.$ And $[x]-1=-1-1=-2.$ Thus, $f(x)=0+\frac{\alpha^{x-1}-2}{x-1}.$ Hence $\lim_{x\to 0^-} f(x)=\lim_{x\to 0^-}\frac{\alpha^{x-1}-2}{x-1}.$ Since the denominator tends to $-1\neq 0$, direct substitution works: $\lim_{x\to 0^-} f(x)=\frac{\alpha^{-1}-2}{-1}=2-\frac1\alpha.$

2. Use the given condition

Given that the left hand limit equals $\alpha-\frac43.$ So, $2-\frac1\alpha=\alpha-\frac43.$ Rearrange: $\frac23-\frac1\alpha=\alpha-2?$ Better solve directly: $2-\frac1\alpha=\alpha-\frac43$ $\frac{10}{3}-\alpha-\frac1\alpha=0.$ Multiply by $3\alpha$: $10\alpha-3\alpha^2-3=0$ $3\alpha^2-10\alpha+3=0.$ Now factor: $3\alpha^2-10\alpha+3=(3\alpha-1)(\alpha-3)=0.$ Thus, $\alpha=\frac13 \quad \text{or} \quad \alpha=3.$

3. Integral value of $\alpha$

The integral value is $\boxed{3}.$

4. Compare with stored answer

Stored correct answer: $3$.

Our derived answer matches the stored answer.