Let $I=\int\dfrac {(12z-7)}{(z-1)^2(2z+3)}$

Circle $|z+1|=\sqrt3$ has centre $(0,-1)$ and radius $\sqrt3$

Here, $Z_0=\dfrac {-3}2$ lies outside while $ Z_0=1$ lies inside the side

$Z_0=1$ in Q pole of order 2

$$\therefore I=\int\dfrac {\frac {(12z-7)}{(2z+3)}}{(z-1)^2}dz$$

$Let \space \space f(z)=\dfrac {12z-7}{2z+3}\space and \space \space Z_0=1\\ \therefore f^1(z)=\dfrac {(2z+3)12-(12\times 1-7)2}{(2z-3)^2}\\ \therefore f^1(z_0)=f^1(1)=\dfrac {2(2(1)+3)12-(12\times 1-7)12}{(2\times 1+3)^2}\\ =\dfrac {50}{25}\\ =2$

By Cauchy’s integral formula,

$$\int\dfrac {f((z))}{(z-z_0)^n}dx=\dfrac {2\pi i}{(n-1)!}f^{n-1}(z_0)$$

$\therefore \int\dfrac {\frac {(12z-7)}{(2z+3)}}{(z-1)^2}dz=\dfrac {2\pi i}{(2-1)!}f^1(z_0)\\ \therefore \int\dfrac {(12z-7)}{(z-1)^2(2z+3)}dz=\dfrac {2\pi i}{1!}\times 2\\ =4\pi i$