Integration without substitution How to i integrate this with out substitutions or Partial fraction decomposition ?
($3x^2$+$2$)/[$x^6$($x^2$+1)]
I've got to : 2/x^6(x^2+1),but after this i haven't been able to eliminate the 2.

You can use $\displaystyle\frac{3x^2+2}{x^6(x^2+1)}=\frac{2(x^2+1)}{x^6(x^2+1)}+\frac{x^2}{x^6(x^2+1)}=\frac{2}{x^6}+\frac{1}{x^4(x^2+1)}$
$\displaystyle=\frac{2}{x^6}+\left(\frac{x^2+1}{x^4(x^2+1)}-\frac{x^2}{x^4(x^2+1)}\right)=\frac{2}{x^6}+\frac{1}{x^4}-\frac{1}{x^2(x^2+1)}$
$\displaystyle=\frac{2}{x^6}+\frac{1}{x^4}-\left(\frac{x^2+1}{x^2(x^2+1)}-\frac{x^2}{x^2(x^2+1)}\right)=\frac{2}{x^6}+\frac{1}{x^4}-\frac{1}{x^2}+\frac{1}{x^2+1}.$
[Notice that this gives the partial fraction decomposition of this particular function.]
