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.

I learned something weird in attempting this integral. When you see $x^2+a^2$, the four substitutions that come to mind are $x=a\sec\theta$, $x=a\csc\theta$, $x=a\sinh\theta$, and $x=a\,\text{csch}\,\theta$. The first $3$ leave a big power of a trigonometric or hyperbolic function in the denominator, so I tried $x=\text{csch}\theta$.
$$\begin{align}\int\frac{x^2+3}{x^6(x^2+1)}dx&=-\int\frac{(\text{csch}^2\theta+3)\text{csch}\theta\coth\theta}{\text{csch}^2\theta\coth^2\theta}d\theta\\
&=-\int\frac{(1+3\sinh^2\theta)\sinh^4\theta}{\cosh\theta}d\theta\\
&=\int\left(-3\cosh\theta\sinh^4\theta+2\cosh\theta\sinh^2\theta-2\cosh\theta+2\text{sech}\theta\right)d\theta\end{align}$$
Now, the first $3$ terms above are no problem but the last one has the issue that the other $3$ secants have a common form for their indefinite integrals,
$$\int\sec\theta d\theta=\ln|\sec\theta+\tan\theta|+C$$
$$\int\csc\theta d\theta=-\ln|\csc\theta+\cot\theta|+C$$
$$\int\text{csch}\,\theta d\theta=-\ln|\text{csch}\,\theta+\coth\theta|+C$$
But $\text{sech}\,\theta$ doesn't fit this pattern so I can never remember its integral, so I always have to substitute $z=\tanh\left(\frac{\theta}2\right)$ and end up with
$$\int2\text{sech}\,\theta\,d\theta=4\tan^{-1}\left(\tanh\frac{\theta}2\right)+C=4\tan^{-1}\left(\frac1{x+\sqrt{x^2+1}}\right)+C$$
Now, not only does this look horrible, but it exposes the integrator for having used the unimaginative 'Weierstrass' substitution. This can be fixed up because
$$\begin{align}4\tan^{-1}\left(\frac1{x+\sqrt{x^2+1}}\right)&=2\tan^{-1}\left(\frac{\frac2{x+\sqrt{x^2+1}}}{1-\frac1{(x+\sqrt{x^2+1})^2}}\right)\\
&=2\tan^{-1}\left(\frac{2(x+\sqrt{x^2+1})}{x^2+2x\sqrt{x^2+1}+x^2}\right)\\
&=2\tan^{-1}\frac1x=\pi-2\tan^{-1}x\end{align}$$
So that means that
$$\int2\text{sech}\theta\,d\theta=-2\tan^{-1}x+C$$
But that in turn implies that
$$\int\text{sech}\,\theta\,d\theta=\tan^{-1}\sinh\theta+C_1=-\tan^{-1}\text{csch}\,\theta+C_2$$
OK, so I will have to remember that. Differentiation confirms this identity.
Maybe it would be easier to remember the forms
$$\int\text{csch}\,\theta\,d\theta=-\coth^{-1}\cosh\theta+C$$
$$\int\sec\theta\,d\theta=\tanh^{-1}\sin\theta+C$$
$$\int\csc\theta\,d\theta=-\tanh^{-1}\cos\theta+C$$
Nahh... Getting back to the task at hand,
$$\begin{align}\int\frac{x^2+3}{x^6(x^2+1)}dx&=-\frac35\sinh^5\theta+\frac23\sinh^3\theta-2\sinh\theta-2\tan^{-1}\text{csch}\theta+C\\
&=-\frac3{5x^5}+\frac2{3x^3}-\frac2x-2\tan^{-1}x+C\end{align}$$
