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.
 A: $$\frac{x^2+3}{x^6(x^2+1)}=\frac{2x^4-2x^2+3}{x^6}-\frac{2}{x^2+1}=2x^{-2}-2x^{-4}+3x^{-6}-2\cdot\frac{1}{x^2+1}.$$
A: 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.
A: Once you have the expression equal to $\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)}$ (as you stated in the question), you can get to the partial fraction decomposition without partial fractions. Using the formula for an infinite geometric series, expand $\frac{1}{x^2+1}=1-x^2+x^4-x^6\dots$, which allows you to write your expression as
$$\frac{1}{x^6}+\frac{2}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-2(1-x^2+x^4-x^6\dots)=\frac{3}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-\frac{2}{1+x^2}.$$
A: $\bf{My\; Solution}$ 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 $\displaystyle \displaystyle \int\frac{x^2+3}{x^6(x^2+1)}dx = -\frac{1}{5x^5}-\frac{2}{5x^5}+\frac{2}{3x^3}-\frac{2}{2x^2}+2\tan^{-1}\left(\frac{1}{x}\right)+\mathcal{C}$
