Show that $ \int_{-\infty}^{\infty} \frac{x^3}{(x^2+4)(x^2+1)}\, dx$ does not converge I noticed that $\displaystyle \int_{-a}^{b} \frac{x^3}{(x^2+4)(x^2+1)}$ will converge to $0$ whenever $a=b$ and will converge to some value whenever $a,b$ are in the reals (excluding infinity). How would you show that this integral does not converge over the interval $(-\infty, \infty)$ to someone who is not convinced? 
When I evaluated this integral using complex residues for the interval $(-\infty, \infty)$ $(Res[z=2i] + Res[z=i])$,I got a complex value as an answer. 
 A: $\displaystyle \frac{x^3}{(x^2+4)(x^2+1)} \geq \frac{x^3}{4x^4}=\frac{1}{4x}$
for sufficiently large $x$.
OBS: And what do you mean by "will converge to $0$ whenever $a=b$ and will converge to some value whenever a,b are in the reals (excluding infinity)"? The integral on those values is a bona fide, well-defined, riemann-integral. There is no "convergence". 
A: Beside the simple point from Aloizio Macedo which would apply to many cases, may be, going further, you could use the fact that $$\frac{x^3}{(x^2+4)(x^2+1)}=\frac{4 x}{3 \left(x^2+4\right)}-\frac{x}{3 \left(x^2+1\right)}=\frac 23 \frac{2 x}{x^2+4}-\frac 16 \frac{2 x}{x^2+1}$$ So $$\int\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\frac{2}{3} \log \left(x^2+4\right)-\frac{1}{6} \log \left(x^2+1\right)$$ Already, the doefficients show what would be the problem $\frac 23 -\frac 16=+\frac 12$.
Since you already noticed that, if $a=b$, the result is $0$  then $$\int_{-a}^b\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\int_{a}^b\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\frac{1}{6} \log \Big(\frac{(b^2+4)^4\,(a^2+1)}{(a^2+4)^4\,(b^2+1)}\Big)$$ which, forgetting a constant which depend on $a$, is basically $\log(b)$  for large values of $b$.
