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.

If you don't want to go the formal partial fractions route, you can systematically chip away at the denominator as follows:
$$\begin{align}
{x^2+3\over x^6(x^2+1)}&={3(x^2+1)-2x^2\over x^6(x^2+1)}\\
&={3\over x^6}-{2\over x^4(x^2+1)}\\
&={3\over x^6}-{2(x^2+1)-2x^2\over x^4(x^2+1)}\\
&={3\over x^6}-{2\over x^4}+{2\over x^2(x^2+1)}\\
&={3\over x^6}-{2\over x^4}+{2(x^2+1)-2x^2\over x^2(x^2+1)}\\
&={3\over x^6}-{2\over x^4}+{2\over x^2}-{2\over x^2+1}\\
\end{align}$$
The final batch of terms are all easy to integrate, provided you recognize $1/(x^2+1)$ as the derivative of $\arctan x$.
Note:  A solid understanding of partial fractions lets you write
$${x^2+3\over x^6(x^2+1)}={Ax^4+Bx^2+C\over x^6}+{D\over x^2+1}$$
(because $x$ appears only to even powers in the expression you're trying to decompose), and then you can solve for the coefficients $A$, $B$, $C$, and $D$.  You get the same answer.
