If $\int_1^ \infty \frac {x^3+3}{x^6(x^2+1)} \, \mathrm d x=\frac{a+b\pi}{c}$, then find $a,b,c$. 
If
$$\int_1^ \infty \frac {x^3+3}{x^6(x^2+1)} \, \mathrm d x=\frac{a+b\pi}{c} $$
then find $a, b, c$.

Now, using partial fractions I calculated
$$a = 62-10\ln (2) \qquad\qquad b = -15 \qquad\qquad c = 20$$
but it took me more than 45 minutes to do all the work. The question was asked in an MCQ exam where only 4-5 minutes are available. I am probably missing something which can help to solve it. Thanks!
Note it was asked in an exam for students of grade 12 so I basically don't know very complex integrations thus I am searching for integration via elementary functions only.

$$\frac {x^3+3}{x^6(x^2+1)}=\frac{3}{x^6}-\frac{3}{x^4}+\frac{1}{x^3}+\frac{3}{
   x^2}-\frac{1}{x}+\frac{x}{x^2+1}-\frac{3}{x^2+1}$$ So $$\int\frac {x^3+3}{x^6(x^2+1)}\,dx=-\frac{3}{5 x^5}+\frac{1}{x^3}-\frac{1}{2 x^2}+\frac{1}{2} \log
   \left(x^2+1\right)-\frac{3}{x}-\log (x)-3 \tan ^{-1}(x)$$ that is to say $$-\frac{3}{5 x^5}+\frac{1}{x^3}-\frac{1}{2 x^2}-\frac{3}{x}+\frac{1}{2} \log
   \left(\frac{x^2+1}{x^2}\right)-3 \tan ^{-1}(x)$$ Now, using the bounds, it seems to be quite fast.
