The function
\[
f(x)=\frac{x-4}{x^{2}+8 x}
\]
is negative on $(3,4)$ and positive on $(4,5)$. Find the area of the region bounded by $f(x)$, the $x$-axis, and the vertical lines $x=3$ and $x=5$. Round to 2 decimal places.
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Step 1 ：Define the function \(f(x) = \frac{x-4}{x^{2} + 8x}\).

Step 2 ：Calculate the integral of the function from 3 to 4. The absolute value is taken because the function is negative in this interval.

Step 3 ：The integral from 3 to 4 is approximately 0.013323970741445814.

Step 4 ：Calculate the integral of the function from 4 to 5. The function is positive in this interval, so the absolute value is not necessary.

Step 5 ：The integral from 4 to 5 is approximately 0.008492285853199761.

Step 6 ：Add the two areas to get the total area of the region bounded by the function, the x-axis, and the vertical lines \(x=3\) and \(x=5\).

Step 7 ：The total area is approximately 0.02.

Step 8 ：Final Answer: The area of the region bounded by the function \(f(x)\), the x-axis, and the vertical lines \(x=3\) and \(x=5\) is \(\boxed{0.02}\).