<s>[INST] 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?
 [/INST] For $x\in (-2,-1)$ we get
\begin{align*}
3^x-4^x&>0\\
\ln(x+2)&<0\\
x^2-3x-4=\left(x+1\right)\left(x-4\right)&>0
\end{align*}
Then
$$x\in (-2,-1)\quad\implies \quad \frac{(3^x-4^2)\ln(x+2)}{x^2-3x-4}<0$$
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