# Math Help - integral

1. ## integral

integral((x^2)-1/((x^2)+6x+9)

2. Originally Posted by lord12
integral((x^2)-1/((x^2)+6x+9)
I assume this is

$\int\frac{x^2-1}{x^2+6x+9}dx$

Well $\frac{x^2-1}{x^2+6x+9}=1-\frac{6}{x+3}+\frac{8}{(x+3)^2}$

3. Edit: eep sorry again mathstud. Please disregeard this post

I think you mean
$\int\frac{x^2-1}{x^2+6x+9}dx$
You have an extra open bracket that you haven't closed at the start of $x^2$ which makes thins confusing.

First we need to reduce the numerator to a degree smaller than the denominator using long division. Here is an example of how to do it using 2 common notations. you should get
$\int 1+\frac{-6x-10}{x^2+6x+9}dx$
Do you know how to proceed from here?

$\int\frac{x^2-1}{x^2+6x+9}dx$
You have an extra open bracket that you haven't closed at the start of $x^2$ which makes thins confusing.
$\int 1+\frac{-6x-10}{x^2+6x+9}dx$