$$\int_{-\infty}^{\infty}\frac{dx}{(x^2+2)^3}$$
I know that I can use a complex function $f(z)$ and I must deal with: $$\int_{-\infty}^{\infty}\frac{dz}{(z^2+2)^3}$$
So I need the roots of $z$
$$z_0 = i\sqrt2, z_1 = - i \sqrt 2$$
And only work with $z_0$
$$Res(f(z),z_0) = \frac{1}{2} \lim_{z \to i\sqrt2}\frac{d^2}{dz^2}\left[(z-i\sqrt 2)^3 \frac{dz}{(z^2 + 2)^3}\right]$$
Is this the way to go?
In addition to the other answer.
You can also Let
$$I(b)=\int \frac{1}{x^2+b} dx$$
Integrate. Using the substitution $x=\sqrt{b} \tan (\theta)$. Then for $b>0$:
$$I(b)=\frac{\arctan (\frac{x}{\sqrt{b}})}{\sqrt{b}}+C$$
Find
$$\frac{I''(b)}{2}+C_2=\int \frac{1}{(x^2+b)^3} dx$$
Well yes , that is the way to go but you've made mistakes when evaluating the poles. I would write it a little bit different. Consider the function $\displaystyle f(z)=\frac{1}{(z^2+3)^3}$. In the upper half plane we you have only one pole , namely $z=i\sqrt{3}$ of order $3$. The residue there is
$$\mathfrak{Res}\left ( f; i \sqrt{3} \right )= \frac{1}{2}\lim_{z \rightarrow i \sqrt{3}} \frac{\mathrm{d^2} }{\mathrm{d} z^2} \left ( z-i \sqrt{3} \right )^3 f(z)= -\frac{i}{48 \sqrt{3}}$$
Now choose a semicircle as you contour in the upper half plane. The contour integral of $f$ is
$$\oint \limits_{C} f(z) \, {\rm d}z = -2\pi i \cdot \frac{i}{48 \sqrt{3}}= \frac{\pi}{24 \sqrt{3}}$$
The integral of the arc vanishes. (leave the details to you). Thus:
$$\int_{-\infty}^{\infty} \frac{{\rm d}x}{\left ( x^2+3 \right )^3} = \frac{\pi}{24 \sqrt{3}}$$
Expanding about the pole at $x=\sqrt2i$, and using the simplification $z=x-\sqrt2i$, we get $$ \begin{align} \frac1{\left(x^2+2\right)^3} &=\left(\left(z+\sqrt2i\right)^2+2\right)^{-3}\\ &=\left(z^2+\sqrt8i\,z\right)^{-3}\\ &=\frac1{(\sqrt8i)^3}z^{-3}\left(1+\frac1{\sqrt8i}z\right)^{-3} \end{align} $$ Using the Binomial Theorem, the coefficient of $z^{-1}$ is therefore $$ \frac1{(\sqrt8i)^5}\binom{-3}{2}=\frac{6}{(\sqrt8i)^5}=\frac{3}{64\sqrt2i} $$ Using the contour along the real axis then returning along the infinite circle around the upper half plane, we get $$ \int_{-\infty}^\infty\frac{\mathrm{d}x}{\left(x^2+2\right)^3}=\frac{3\pi}{32\sqrt2} $$
