Help with partial fraction $\int_{4}^{5}\frac{x^3-3x^2-9}{x^3-3x^2}dx$ I can't figure out why I keep getting this answer wrong.  
First, simplifying the answer:
$$\int_{4}^{5}1-\frac{9}{x^2(x-3)}dx$$
Setting up the partial fraction:
$$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-3}$$
Expanding...
$$\frac{A(x^2)(x-3)+B(x)(x-3)+Cx^3}{x^2(x-3)}$$
Further expansion...
$$\frac{A(x^3-3x^2)+B(x^2-3x)+Cx^3}{x^2(x-3)}$$
Grouping...
$$\frac{(A+C)x^3+(B-3A)x^2(-3Bx)}{x^2(x-3)}$$
After solving the system of equations I got the following values:
$$A=1, B=3, C=-1$$
So my new integral is now:
$$\int_{4}^{5}\frac{1}{x}+\frac{3}{x^2}-\frac{1}{x-3}dx$$
Solving I got this:
$$ln|x|+2xln|x^2|-ln|x-3||_{4}^{5}$$
Final answer:
$$ln5+10ln25-ln2-(ln4+8ln16)$$

Recall that the original integrand was $$1 - \frac{9}{x^2(x-3)}$$ Did you forget to account for the $1$, which would integrate to $x$, and/or forget that you are subtracting the quotient from $1$?
So you should end with $$\int_{4}^{5} 1 - \Big(\frac{1}{x}+\frac{3}{x^2}-\frac{1}{x-3}\Big)dx$$
Also, use the power rule for integrating $$\int \frac 3{x^2} \,dx = -3x^{-1} = -\frac 3x$$
