Find $\int_\Gamma\frac{2z+j}{z^3(z^2+1)}\mathrm{d}z$ where $Γ:|z-1-i| = 2$ pls, some ideas for integral solution (residue theory)?
$$\int_\Gamma\dfrac{2z+j}{z^3(z^2+1)}\mathrm{d}z$$
Where $Γ:|z-1-i| = 2$ is positively oriented circle.
Thx, for help!

So, my solution is:
Is it correct ??
pole z1 = 0 ( order 3 pole );
pole z2 = -i ( simple pole );
pole z3 = i ( simple pole );
z1 : |0-1-i| = sqrt(2) < 2 => in circle;
z2 : |-i-1-i| = sqrt(5) > 2 => not in circle;
z3 : |i-1-i| = 1 < 2 => in circle;
$$
\underset{z=0}{res}\frac{2z+i}{z^3(z^2+1)}=-i
$$
$$
\underset{z=i}{res}\frac{2z+i}{z^3(z^2+1)}=\frac{3i}{2}
$$
$$
\int_\Gamma\dfrac{2z+i}{z^3(z^2+1)}\mathrm{d}z = 2\pi i(\underset{z=0}{res} f(z) + \underset{z=i}{res} f(z)) = 2\pi i(-i+\dfrac{3i}{2}) = -\pi 
$$
