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.

Both methods work just fine here:
(1) Partial fractions One can apply the Method of Partial Fractions to any rational expression, and since the denominator of the integrand is already factored into a product of irreducible polynomials (over $\Bbb R$), we can proceed directly to writing the decomposition:
$$\frac{x^2 + 3}{x^6 (x^2 + 1)} = \frac{A}{x^6} + \frac{B}{x^5} + \frac{C}{x^4} + \frac{D}{x^3} + \frac{E}{x^2} + \frac{F}{x} + \frac{G}{x^2 + 1} .$$
In fact, since the integrand is even, only even terms appear in the decomposition: $$\frac{x^2 + 3}{x^6 (x^2 + 1)} = \frac{A}{x^6} + \frac{C}{x^4} + \frac{E}{x^2} + \frac{G}{x^2 + 1} .$$
(2) Trigonometric substitution This is probably less efficient than partial fractions here, but it's not a bad option: The usual substitution $x = \tan \theta, dx = \sec^2 \theta\, d\theta$, transforms the integral to $$\int \frac{\tan^2 \theta + 3}{\tan^6 \theta},$$ which can be handled in several ways. One approach is to write the numerator as $\sec^2 \theta + 2$ and splitting the integral as $$\int \frac{\sec^2 \theta \,d\theta}{\tan^6 \theta} + 2\int \cot^4 \theta \,d\theta .$$ The first can be handled with substitution and the second with the usual trick of writing $\cot^2 = \csc^2 - 1$ and applying the elementary integral $\int \csc^2 \theta = - \cot \theta + C$.
