Logarithmic inequality

Observe that $\frac{\log_{2}(x+1)}{x-1}$ is defined for $x>-1,x≠1$ and also $x$ should not be equal to $0$ as the expression would become $0$ .

Let $n=\log_{2}(x+1)$ and $m=x-1$ .

• For $0>x>-1: n$ is negative and $m$ is also negative, so $\frac{n}{m}>0$ .

• For $1>x>0:n$ is positive but $m$ is negative, so $\frac{n}{m}<0$ .

• For $x>1:n$ is positive and $m$ is also positive, so $\frac{n}{m}>0$ .

Therefore, $x\in (-1,0) \cup (1,\infty)$ .

$\Rightarrow a+b+c=-1+0+1=\boxed{0}$