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

Solution 1:

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

Solution 2:

$$\frac {x^3+3}{x^6(x^2+1)}=\frac{3}{x^6}-\frac{3}{x^4}+\frac{1}{x^3}+\frac{3}{ x^2}-\frac{1}{x}+\frac{x}{x^2+1}-\frac{3}{x^2+1}$$ 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.