Step 1: Substitute to move to positive limits:
Put $x=-t$, so $dx=-dt$ and $\frac{1}{x}=-\frac{1}{t}$.
When $x=-5$, $t=5$. When $x=-1$, $t=1$.
Step 2: Rewrite and integrate:
\[ \int_{-5}^{-1}\frac{dx}{x}=\int_{5}^{1}\left(-\frac{1}{t}\right)(-dt)=\int_{5}^{1}\frac{dt}{t} \]
\[ =\Big[\log t\Big]_{5}^{1}=\log 1-\log 5=-\log 5 \]
Swapping the limits would give \(+\log 5\), so the order of limits matters here.
Step 3: Reason about the options:
The graph of $\frac{1}{x}$ is below the x-axis between -5 and -1, so the area-type integral must be negative.
\(-\log 5\) is negative, which fits.
Options with $x^{6}$ and $x^{-6}$ depend on x, so they cannot be the value of a definite integral.
$\log(-5)$ does not exist for real numbers.
Final Answer:
The integral equals $-\log 5$, so option (A) is correct. \[ \boxed{-\log 5} \]