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.
 A: $$\frac {x^3+3}{x^6(x^2+1)}=\frac{3}{x^6}-\frac{3}{x^4}+\frac{1}{x^3}+\frac{3}{
   x^2}-\frac{1}{x}+\frac{x}{x^2+1}-\frac{3}{x^2+1}$$ 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.
A: First, break the integral in two
$$ \int_0^{\infty} \frac{ x^3 + 3 }{x^6(x^2+1)} = \int_1^{\infty} \frac{dx}{x^3(x^2+1)} + 3 \int_1^{\infty} \frac{dx}{x^6(x^2+1)} $$
First, we integrat the second integral. Notice 
$$ \int\limits_1^{\infty} \frac{(1 +x^2 - x^2) dx }{x^6(x^2+1)} = \int\limits_1^{\infty} \frac{dx}{x^6} - \int\limits_1^{\infty} \frac{ d x}{x^4(x^2+1)} = \frac{1}{5} - \int\limits_1^{\infty} \frac{dx}{x^4} + \int\limits_1^{\infty} \frac{ dx }{x^2 ( x^2+1)} =$$
$$ = \frac{1}{5} - \frac{1}{3} + \int\limits_1^{\infty} \left( \frac{1}{x^2} - \frac{1}{1+x^2} \right) dx = -\frac{2}{15} + \frac{ \pi }{2} +1 $$
Now, for the first integral, use same trick
$$ \int\limits_1^{\infty} \frac{ (1 + x^2 - x^2)dx}{x^3(x^2+1)} = \int_1^{\infty} \frac{dx}{x^3} - \int_1^{\infty} \frac{ dx }{x(x^2+1)} = \frac{1}{2} - \int_1^{\infty} \frac{ dx }{x(x^2+1)}$$
Using $x = \tan t$, we solve the last integral easily,
$$ \int_1^{\infty} \frac{ \sec^2 t dt }{tan t \sec^2 t } = \int_1^{\infty} \frac{ \cos t dt }{\sin t} = \ln ( \sin t ) = \ln ( \frac{ x }{\sqrt{x^2+1}}) = \frac{1}{2} \ln \left( \frac{x^2 }{x^2+1} \right) \bigg|_1^{\infty} = \frac{1}{2} \ln (1/2)$$
Thus,
$$ \int_0^{\infty} \frac{ x^3 + 3 }{x^6(x^2+1)} = \frac{-2}{15} + \frac{\pi}{2} + 1 + \frac{1}{2} - \frac{1}{2} \ln(1/2) = \boxed{ \frac{30 \ln 2+ 81 + 30 \pi }{60}}$$
A: Use the change of variable $x\mapsto x^{-1}$ to turn your integral into
$$
\int_0^1\frac{x^3(1+3x^3)}{1+x^2}\,dx=\int_0^1\frac{x^3}{1+x^2}\,dx+3\int_0^1\frac{(1+x^2-1)^3}{1+x^2}\,dx\\
=\frac{1}{2}\int_0^1\frac{x\,dx}{1+x}+3\int_0^1(1+x^2)^2\,dx-9\int_0^1(1+x^2)\,dx+9\int_0^1\,dx-3\int_0^1\frac{\,dx}{1+x^2}\\
=\frac{1}{2}\int_1^2\frac{x-1}{x}\,dx+3\int_0^1\,dx-3\int_0^1x^2\,dx+3\int_0^1x^4\,dx-3\int_0^1\frac{\,dx}{1+x^2}
$$
