<s>[INST] Evaluate the Integral: $\int_{4}^{5}\frac{x^3-3x^2-9}{x^3-3x^2}$ $$\int_{4}^{5}\frac{x^3-3x^2-9}{x^3-3x^2}$$
I used long division to get:
$$\int_{4}^{5}1+\frac{9}{x^3-3x^2}$$
Factored the denominator:
$$x^2(x-3)$$
Broke the rational function into partial fractions:
$$\frac{9}{x^2(x-3)} = \frac{A}{x-3} + \frac{B}{x} + \frac{C}{x^2}$$
Solved for A, B, and C and got:
$$A = 1,\quad B = -1,\quad
C = -3$$
Substituted the values into the equation:
$$\int_{4}^{5}1 + \frac{1}{x-3} - \frac{1}{x} - \frac{3}{x^2}$$
Found the anti-derivative:
$$[x + \ln|x-3| - \ln|x| - 3\ln|x^2|]_{4}^{5}$$
I keep getting the answer wrong so at this point, I don't know what I'm doing wrong. Any help would be appreciated.
 [/INST] Your only errors are a trivial one, 
$$\frac{x^3-3x^2 -9}{x^3-3x^2} = 1 - \frac{9}{x^3-3x^2}$$
(you have a $+$ instead), and a somewhat more serious one that lies 
in evaluating the integral of $\frac{3}{x^2}$. The latter is not $3\ln(x^2)$.
First, remember that the integral
$$\int\frac{dx}{f(x)}$$
is almost never equal to $\ln|f(x)|$. And second, 
remember that 
$$\int x^n\,dx = \left\{\begin{array}{ll}
\frac{1}{n+1}x^{n+1} +C &\text{if }n\neq -1;\\
\strut\\
\ln|x| + C &\text{if }n=-1.
\end{array}\right.$$
So 
$$\begin{align*}
\int_4^5 -\frac{3}{x^2}\,dx &= -3\int_4^5 x^{-2}\,dx\\
 &= -3\left(\left. \frac{1}{-2+1}x^{-2+1}\right|_{4}^5\right)\\
 &= -3\left(\left.\frac{1}{-1}x^-1\right|_{4}^5\right)\\
&= -3\left(\left. -\frac{1}{x}\right|_{4}^5\right)\\
&= -3\left( -\frac{1}{5} + \frac{1}{4}\right).
\end{align*}$$
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