$$ \int_{-\infty}^{+\infty}\frac{\sin(x)}{2x^{5}-3jx^{3}+2x}dx=0 $$
$$(1)\int_{-\infty}^{+\infty}\frac{\sin(x)}{2x^{5}-3jx^{3}+2x}dx=\frac{1}{2j}(\int_{-\infty}^{+\infty}\frac{e^{jx}}{2x^{5}-3jx^{3}+2x}dx-\int_{-\infty}^{+\infty}\frac{e^{-jx}}{2x^{5}-3jx^{3}+2x}dx) $$
Singularities:
$z_0=0\,,\, z_1= 1+j\,,\,z_2=-1-j\,,\,z_3 = \frac{1}{2}-\frac{1}{2}j\,,\,z_4 = -\frac{1}{2}+\frac{1}{2}j$
They are all simple poles.
$$f(z)=\frac{e^{jz}}{2z^{5}-3jz^{3}+2z}$$ $$g(z)=\frac{e^{-jz}}{2z^{5}-3jz^{3}+2z}$$
Residues of $f(z)$ e $g(z)$: $$Res_f[0]=Res_g[0]=\frac{1}{2}$$ $$Res_f[1+j]=Res_g[-1-j]=\frac{e^{-1+j}}{-20}$$ $$Res_f[-\frac{1}{2}+\frac{1}{2}j]=Res_g[\frac{1}{2}-\frac{1}{2}j]=\frac{-e^{-\frac{1}{2}-\frac{1}{2}j}}{5}$$
Therefore,
$$\int_{-\infty}^{+\infty}\frac{\sin(x)}{2x^{5}-3jx^{3}+2x}dx=\frac{1}{2j}2\Pi j[0]=0$$
Edit:
$$\int_{-\infty}^{+\infty}\frac{\sin(x)}{2x^{5}-3jx^{3}+2x}dx=\Pi(1+\frac{e^{-1+j}}{10}-\frac{2e^{-\frac{1}{2}-\frac{1}{2}j}}{5})$$
Right?