$\displaystyle
\frac{x - 3}{x^2 + 1}
$
This looks like a tan^-1x type answer but the actual answer is different and I am not sure of the process to get there.
okay so separate it out:
$\displaystyle \frac{ x-3}{x^{2}+1 } = \frac{x }{ x^2+1} - \frac{ 3}{x^2+1 }$
then
$\displaystyle
\int_{ }^{ } \frac{ x-3}{x^{2}+1 }dx = \int_{ }^{ } \frac{x }{ x^2+1}dx - \int_{ }^{ } \frac{ 3}{x^2+1 }dx$
The part with the 3 is just arctan. Use a substitution on the other part:
u = x^2 + 1 --> du = 2xdx ---> dx = du/(2x), and so the integral should give a natural log.