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.

Use the change of variable $x\mapsto x^{-1}$ to turn your integral into
$$
\int_0^1\frac{x^3(1+3x^3)}{1+x^2}\,dx=\int_0^1\frac{x^3}{1+x^2}\,dx+3\int_0^1\frac{(1+x^2-1)^3}{1+x^2}\,dx\\
=\frac{1}{2}\int_0^1\frac{x\,dx}{1+x}+3\int_0^1(1+x^2)^2\,dx-9\int_0^1(1+x^2)\,dx+9\int_0^1\,dx-3\int_0^1\frac{\,dx}{1+x^2}\\
=\frac{1}{2}\int_1^2\frac{x-1}{x}\,dx+3\int_0^1\,dx-3\int_0^1x^2\,dx+3\int_0^1x^4\,dx-3\int_0^1\frac{\,dx}{1+x^2}
$$
