Does this open mouth make any Difference?

Just find the critical points. they are $0~and~1$

solve using wavy curve u get $(- \infty,0)$ and $(1,\infty)$

Also the argument of logarithm i.e. $x+1>0$

taking the intersection of all values, the answer is $(-1,0)~U~(1,\infty)$

I know it is a long solution and it can be done in a shorter way, but since the shorter way has already been posted, I have posted the long algebraic approach.

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Let's take 3 cases.

Case 1: x+1>1 ( $\therefore x\in \left( 0,\infty \right)$ )

$\frac { \log _{ 2 }{ \left( x+1 \right) } }{ x-1 } >0$

Since numerator of LHS is positive, the denominator should be positive too.

$\therefore$ x-1>0

$\therefore$ x>1 ( $\therefore x\in \left( 1,\infty \right)$ )

$x\in \left( 0,\infty \right) \cap \left( 1,\infty \right) \\ \Rightarrow x\in \left( 1,\infty \right)$ for x+1>1

Case 2: x+1<1 ( $\therefore x\in \left( -\infty ,0 \right)$ )

$\frac { \log _{ 2 }{ \left( x+1 \right) } }{ x-1 } >0$

Since numerator of LHS is negative, the denominator should be negative too.

$\therefore$ x-1<0

$\therefore$ x<1 ( $\therefore x\in \left( -\infty,1 \right)$ )

$x\in \left( -\infty ,0 \right) \cap \left( -\infty ,1 \right) \\ \Rightarrow x\in \left( -\infty ,0 \right)$ for x+1<1

Case 3: x+1=0

$\frac { \log _{ 2 }{ \left( x+1 \right) } }{ x-1 } >0$

$\frac { \log _{ 2 }{ \left( 1\right) } }{ x-1 } >0$

$0 >0$

$\therefore x\in \phi$ for x+1=0

Taking union of all cases:

$x\in \left( -\infty ,0 \right) \cup \left( 1,\infty \right)$

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Now we look at the conditions.

Condition 1: x+1>0 (because it is in the logarithm)

x+1>0

$\therefore$ x>-1

$\therefore x\in \left( -1,\infty \right)$

Condition 2: x-1 $\neq$ 0

$\\ \therefore x\neq 1$

$\therefore \left( -\infty ,1 \right) \cup \left( 1,\infty \right)$

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Intersecting cases and conditions, we get:

$\therefore x\in \left[ \left( -\infty ,0 \right) \cup \left( 1,\infty \right) \right] \cap \left( -1,\infty \right) \cap \left[ \left( -\infty ,1 \right) \cup \left( 1,\infty \right) \right]$

$\therefore x\in \left( -1,0 \right) \cup \left( 1,\infty \right)$

So, -1 is the incorrect solution. But if we ignore the $\left( 1,\infty \right)$ part, we get-

$-1+0=\boxed { -1 }$

What I did is this.

Case 1: numerator, $p(x)$ and denominator, $q(x)$ such that $p(x) > 0, q(x) > 0$

$p(x) > 0 \Rightarrow log_2 (x + 1) > 0 \Rightarrow x > 0$ or $x \in (0, \infty)$ $q(x) > 0 \Rightarrow x - 1 > 0 \Rightarrow x > 1$ or $x \in (1, \infty)$

Hence $x \in (1, \infty) \cap (0, \infty) \Rightarrow x \in (1, \infty)$

Here's the fun part:

Case 2: numerator, $p(x)$ and denominator, $q(x)$ such that $p(x) < 0, q(x) < 0$

$p(x) < 0 \Rightarrow log_2 (x + 1) < 0 \Rightarrow x \in (-1, 0)$ $q(x) > 0 \Rightarrow x - 1 < 0 \Rightarrow x < 1$ or $x \in (- \infty, 1)$

Hence $x \in (-1, 0) \cap (- \infty, 1) \Rightarrow x \in (-1, 0)$

Note that $(-1, 0) = (a, b)$ , therefore $a + b = -1 + 0 = -1$

We are done!