Show that $\int_{-\infty}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$ is convergent. 
Show that $$\int_{-\infty}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent.

$$\int_{0}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent by comparison test with $\int_{0}^{\infty} \dfrac{1}{x^4} dx$.
How do i show that $$\int_{-\infty}^{0} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent?

Observe that for every $x\in\mathbb{R}$ you have:
$$1\geq 0 \Longrightarrow x^2+1\geq x^2\Longrightarrow \frac{x^{2}}{x^2+1}\leq 1$$
$$x^2\geq 0 \Longrightarrow 1+x^2\geq 1\Longrightarrow \frac{1}{x^2+1}\leq 1$$
multiplying both inequalities together we get
$$\frac{x^2}{(x^2+1)^2}\leq 1$$
this implies ( by multiplying both sides by ${\frac{1}{(x^2+2x+2)}}$):
$$0\leq \frac{x^2}{(x^2+1)^2(x^2+2x+2)}\leq \frac{1}{(x^2+2x+2)}=\frac{1}{(x+1)^2+1}$$
because $x^2+2x+2=(x+1)^2+1>0$
and $$\int_{\mathbb{R}}{\frac{1}{(x+1)^2+1}}dx=\pi$$
