Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $ 
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$

I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.

hint: $\dfrac{x^2+3}{x^6(x^2+1)}= x^{-6}+\dfrac{2}{x^6(x^2+1)}$. At this point, let $x = \dfrac{1}{y}$, then the second part becomes $\dfrac{y^6}{y^2+1}= y^4-\dfrac{y^4}{y^2+1}=y^4-y^2 + \dfrac{y^2}{y^2+1}=y^4-y^2+1 - \dfrac{1}{y^2+1}$,and you can take it from here.
