How do I start with this one?
$\displaystyle
\int \frac {5x^{2}-3} {x^2-9}
$
Since nominator and denominator are x^2 I have to divide them, right?
I am not sure how to do that.
Thanks for any help.
You could do it this way:
$\displaystyle \int \frac {5x^{2}-3} {x^2-9} dx =$
$\displaystyle \int \frac {5x^{2}-45 + 42} {x^2-9} dx =$
$\displaystyle \int \frac {5(x^2 - 9)}{x^2 - 9} dx + \int \frac {42}{x^2 - 9} dx= $
$\displaystyle \int 5 dx + \int \frac {42}{x^2 - 9} dx$
Then use an appropriate substitution on the second integral.