Ignore the one for now, just try to decompose
\[ \frac{9}{x^2(x-3)} =\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-3} \]
Multiply through by x^2(x-3) to get
\[ 9 = Ax(x-3)+ B(x-3) + Cx^2 \]
There is a tricky way, which is faster, or a straightforward way, which takes longer. Which would you prefer?
I think you decompose it to\[\frac{ A }{ x-3 }+\frac{ B+Cx }{ x^2 }\]..not a 100% sure..but there is something different done when you are over a power greater than one. Side note, were did you get A/x from?
@cherio12, when decomposing (something)/x^2 you have
\[\frac{A}{x}+\frac{B}{x^2}\]
I'm not familiar with the formal explanation as to why this happens, but I know if you have factors with multiplicity greater than 1 they're treated as normal linear factors.
Your answer doesn't look wrong at first glance, but I'm still working back. It could be that the program your class uses is very picky about the answer it accepts.
Okay, so using the longer method mentioned by @Jemurray3, I get different values of A, B, and C:
\[1 + \int_{4}^{5}-\frac{9}{x^2(x-3)}dx\]
Focusing on the integral: Decomposing into partial fractions, you get
\[-9=Ax(x-3)+B(x-3)+Cx^2\\
-9=(A+C)x^2+(-3A+B)x-3B\\
\text{Matching up coefficients, you get}\\
\begin{cases}A+C=0\\
-3A+B=0\\
-3B=-9\end{cases}\\
\text{Solving, you get}\\
B=3, A=1, C=-1\]
The method I used works pretty much every time. You're less prone to making mistakes, I think. I'll check your work again; if the answer really is wrong, it'll probably be an algebra mistake.