$$\int_c \frac{z}{z^2 + 4z + 3}dz$$
where C is a circle centered at -1 with radius 2.
$$\int_c \frac{z}{z^2 + 4z + 3}dz = \int_c \frac{z}{(z+3)(z+1)}dz$$
So both singularities z = -1 and z = -3 but z = -3 is on the circle. So can we use Cauchy's formula later on?
Using partial fractions:
$$\int_c \frac{z}{(z+3)(z+1)}dz = \frac{A}{z+3} + \frac{B}{z+1} : z = A(z+1) + B(z+3)$$ and when z = -1, -1 = 2B => $B = \frac{-1}{2} $ and when z = -3 : -2A = -3 so A = $\frac{2}{3}$
so we get:
$$\frac{3}{2} \frac{1}{z+3}dz + \frac{-1}{2} \frac{1}{z+1}dz$$
so if we apply Cauchy's integral formula to both:
$$\frac{3}{2} 2 \pi i (1) + \frac{-1}{2} 2 \pi i (1) = 3 \pi i + - \pi i = 2 \pi i$$
But is that right?