Solving $\dfrac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\leq 0$ 
The solution set of the inequality $\dfrac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\leq 0$ is.

My attempts:
For $x>0\implies 3^x-4^x<0$
$\implies \dfrac{\ln (x+2)}{x^2-3x-4}\geq 0$. Now root of numerator is $x=-1$ and of denominator is $x=4,-1$
By wavy curve method:
$\implies x\epsilon(4,\infty)$
Now for $x\leq 0\implies 3^x-4^x\geq 0$
Hence:
$\dfrac{\ln (x+2)}{x^2-3x-4}\leq 0$
$\implies x\epsilon(-1,0]$
My answer $(-1,0]\cup (4,\infty)$
But book's answer said I missed one interval, please help.
I observed something that $-1$ are common roots in both numerator and denominator, so should we put $-ve$ on LHS of $-1$? If we do this then my answer will be correct, but don't know whether it's correct thinking or not, if it is then why?

The term $3^x-4^x$ is positive for $x<0$ and negative for $x>0$.
The term $\ln(x+2)$ is positive for $x>-1$ and negative for $-2<x<-1$ (undefined for $x\le-2$).
The term $x^2-3x-4$ is positive for $x<-1$ or $x>4$ and negative for $-1<x<4$.
$$
\begin{array}{l|ccccccccc}
&& -2 & & -1 && 0 && 4 \\
\hline
3^x-4^x & <0 &<0 & <0 & <0 & <0 & =0 & >0 & >0 & >0 \\
\ln(x+2) & \text{u} & \text{u} & <0 & =0 & >0 & >0 & >0 & >0 & >0 \\
x^2-3x-4 & >0 & >0 & >0 & =0 & <0 & <0 & <0 & =0 & >0 \\
\hline
f(x) & \text{u} & \text{u} & >0 & \text{u} & >0 & =0 & <0 & \text{u} & >0
\end{array}
$$
where “u” means “undefined”.
The last row is obtained by using the “rule of signs” on the factors. The function is undefined where the denominator vanishes.
Hence you get, for the solutions, $(-2,-1)\cup(-1,0]\cup(4,\infty)$.
The problem with your solution is that two factors vanish at $-1$, so the global sign doesn't change when we cross it.
