$\int_{4}^{5}\frac{x^3-3x^2-9}{x^3-3x^2}$
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.