Hi I'm trying to solve this Partial Fraction Question: I have found A,B, and C. I just cannot get the correct answer.

$\displaystyle \int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx$

$\displaystyle A = -\frac{75}{25}$

$\displaystyle B = 0$

$\displaystyle C = 1$

$\displaystyle \int -\frac{\frac{75}{25}}{x-4} + \int \frac{1}{x^2+9}$

Answer:

$\displaystyle -3ln(x-4)+\frac{1}{2}ln(x^2+9)+C$