# Math Help - Integration by partial fractions

1. ## Integration by partial fractions

It was easier to just make a scan, so here it is:
http://img195.imageshack.us/img195/9144/scan0001tv.jpg

The answer I'm getting is wrong. It's an easy problem, but I'm sure I'm doing something stupid. Please advise.

2. Did you consider splitting the numerator in the first place! 1 - 9/(x^3 - 3x^2) should be a bit easier...

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

$= 1 - \frac{9}{x^2(x-3)}$

$= 1 - [ \frac{1}{x-3} - \frac{x+3}{x^2} ]$

$= 1 + \frac{1}{x} + \frac{3}{x^2} - \frac{1}{x-3}$

Therefore , its integral is

$x + \ln(x) - \frac{3}{x} - \ln(x-3) + C$