Find the value of integral $ \int_{ \mid z \mid =1} \frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } $ Integral to get the result?

$$ \int_{ \mid z \mid =1}  \frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) }  $$

with Solutions ,tnx.

With partial fraction decomposition you have :
$$f(z)=\frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } = 2\frac{1}{z-\frac{1}{3}}+\frac{1}{2}\frac{1}{z-\frac{1}{2}}+\frac{1}{2}\frac{1}{(z-\frac{1}{2})^2}$$
We will try to use Residue theorem now :
$Res(f,\frac{1}{3})=Res(2\frac{1}{z-\frac{1}{3}})=2$
you should also find $Res(f,\frac{1}{2})=\frac{1}{2}$ (you have to justify a bit this calculation).
So finally : $$\int_{ \mid z \mid =1}  \frac{ 30z^{2}-23z+5 }{ (2z-1)^{2}(3z-1) } dz=2\pi i(2+\frac{1}{2})=\pi i5$$
