<s>[INST] Finding the value of integers of a logarithmic function. Expressing $\frac{2x^2-9x-6}{x(x^2-x-6)}$ in partial fraction would give you: $\frac{16}{5(x+2)}$-$\frac{1}{5(x-3)}$-$\frac{1}{x}$
Given that $\int_{4}^6\frac{2x^2-9x-6}{x(x^2-x-6)}dx=ln\frac{m}{n},$ determine the values for the integers m and n.
 [/INST] It turns out the partial fraction decomposition is incorrect.
It can be verified that instead,
$$\begin{align}\int_4^6 \frac{2x^2 - 9x - 6}{x(x^2 - x - 6)} dx &= \int_4^6\frac{1}{x} + \frac{2}{x + 2} - \frac{1}{x-3}dx\\
&=\left[\ln x + 2 \ln (x + 2) - \ln (x - 3)\right]_4^6\\
&= \ln\frac{6}{4} + 2\ln \frac{8}{6} - \ln\frac{3}{1}\\
&= \ln \frac{3}{2} + \ln\frac{16}{9} - \ln 3\\
&= \ln \frac{8}{9}\end{align}$$
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