Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$ This is a repost of a question i had written incorrectly earlier. 
How do I integrate this without substitutions ?
$$
 \frac{x^2+3}{x^6\left(x^2+1\right)}
$$
I got:
$$
 \frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)},
$$
but wasn't able to eliminate the 2.

$\bf{My\; Solution}$ Given $\displaystyle \int\frac{x^2+3}{x^6(x^2+1)}dx = \int\frac{(x^2+1)+2}{x^6(x^2+1)}dx = \int x^{-6}dx+\int\frac{2}{x^6(x^2+1)}dx$
$\displaystyle =-\frac{1}{5}x^{-5}+I,$ where $\displaystyle I = \int\frac{2}{x^6(x^2+1)}dx$
Now Using $\displaystyle x= \frac{1}{t}$ and $\displaystyle dx = -\frac{1}{t^2}$. So Let $\displaystyle I = -\int \frac{2t^6}{1+t^2}dt  = -\int \frac{2(t^6+1)-2}{1+t^2}dt$
So $\displaystyle I = -2\int \frac{t^6+1}{t^2+1}dt+\int \frac{2}{1+t^2}dt = -2\int (t^4-t^2+1)dt+2\tan^{-1}(t)$
$\displaystyle  = -\frac{2t^5}{5}+\frac{2t^3}{3}-\frac{2t^2}{2}+2\tan^{-1}(t)+\mathcal{C} = -\frac{2}{5x^5}+\frac{2}{3x^3}-\frac{2}{2x^2}+2\tan^{-1}\left(\frac{1}{x}\right)+\mathcal{C}$
So $\displaystyle \displaystyle \int\frac{x^2+3}{x^6(x^2+1)}dx = -\frac{1}{5x^5}-\frac{2}{5x^5}+\frac{2}{3x^3}-\frac{2}{2x^2}+2\tan^{-1}\left(\frac{1}{x}\right)+\mathcal{C}$
