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.

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
$$\frac{1}{x^6}+\frac{2}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-2(1-x^2+x^4-x^6\dots)=\frac{3}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-\frac{2}{1+x^2}.$$
