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.
Thanks for your time.
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.
Thanks for your time.
$\displaystyle \frac{ x^3 - 3x^2 - 9}{ x^3 - 3x^2 } $
$\displaystyle = 1 - \frac{9}{x^2(x-3)}$
$\displaystyle = 1 - [ \frac{1}{x-3} - \frac{x+3}{x^2} ] $
$\displaystyle = 1 + \frac{1}{x} + \frac{3}{x^2} - \frac{1}{x-3} $
Therefore , its integral is
$\displaystyle x + \ln(x) - \frac{3}{x} - \ln(x-3) + C$