<s>[INST] 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.
 [/INST] Below is a nice general way of computing the partial fraction that is a generalization of Heaviside's cover up method to nonlinear denominator factors.
$$\begin{eqnarray} \frac{x^2+3}{(x^2+1)x^6} &=&\ \ \frac{ax+b}{x^2+1}+ \frac{f(x)}{x^6}\quad\text{so clearing denominators yields}\\
 x^2+3\ &=&\ \ (ax+b)x^6\! + (x^2+1) f(x)\quad\text{so evaluaing at $x=i$ then $x = -i$}  \\
 2\ &=& -\!ai - b \\
 2\ &=&  \ \ \ \ ai - b\quad\text{so subtracting the prior two equations yields}\\
 0\ &=& -\!2ai\end{eqnarray}$$
Therefore $\ a=0\ $ so $\ 2 = -b\ $ so $\ b = -2\ $ and, finally, we can compute $\,f(x)\,$ by substituting these $\,a,b\,$ values into the first equation. Then the integral of the decomposition is easy.
See here for further discussion of this method - which often greatly simplifies calculations.
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