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.

Since
$$
\begin{align}
\frac{x^2+3}{x^6(x^2+1)}
&=\frac1{x^6}+\frac2{x^6(x^2+1)}\\
&=\frac1{x^6}+\frac{2(x^6+1)}{x^6(x^2+1)}-\frac2{x^2+1}\\
&=\frac1{x^6}+\frac{2(x^4-x^2+1)}{x^6}-\frac2{x^2+1}\\
&=\frac3{x^6}-\frac2{x^4}+\frac2{x^2}-\frac2{x^2+1}
\end{align}
$$
we have
$$
\int\frac{x^2+3}{x^6(x^2+1)}\,\mathrm{d}x
=C-\frac3{5x^5}+\frac2{3x^3}-\frac2x-2\arctan(x)
$$
