The value of $$\int_C \frac{3 z+1}{z(2 z+1)} d z$$​ where C is the circle |z| = 1 is:

Correct option is B

​$$\begin{aligned}&\text{\textbf{Identify Singularities}} \\&\text{The integrand } f(z) = \frac{3z+1}{z(2z+1)} \text{ has singularities at:} \\&\quad 1.\, z = 0 \\&\quad 2.\, 2z+1 = 0 \implies z = -\frac{1}{2} \\&\text{Both lie inside the contour } C \text{ (unit circle) since } |0| = 0 < 1 \text{ and } |-\frac{1}{2}| = \frac{1}{2} < 1. \\[10pt]&\text{\textbf{Compute Residues}} \\[5pt]&\text{Residue at } z = 0: \\&\quad \operatorname{Res}(f,0) = \lim_{z\to 0} z \cdot \frac{3z+1}{z(2z+1)} = \lim_{z\to 0} \frac{3z+1}{2z+1} = 1 \\[5pt]&\text{Residue at } z = -\frac{1}{2}: \\&\quad \operatorname{Res}\left(f,-\frac{1}{2}\right) = \lim_{z\to -\frac{1}{2}} \left(z+\frac{1}{2}\right) \cdot \frac{3z+1}{z(2z+1)} \\&\quad \text{Note: } 2z+1 = 2\left(z+\frac{1}{2}\right) \\&\quad = \lim_{z\to -\frac{1}{2}} \frac{3z+1}{2z} = \frac{-\frac{3}{2}+1}{-1} = \frac{-\frac{1}{2}}{-1} = \frac{1}{2} \\[10pt]&\text{\textbf{Apply Cauchy's Residue Theorem}} \\&\quad \oint_C \frac{3z+1}{z(2z+1)} dz = 2\pi i \left(\operatorname{Res}(f,0) + \operatorname{Res}\left(f,-\frac{1}{2}\right)\right) \\&\quad = 2\pi i \left(1 + \frac{1}{2}\right) = \boxed{3\pi i}\end{aligned}$$​​