What I have is $\int_{C[0, 3]} \frac{dz}{(z + 4)(z + i)(z - i)}$ and I know that there are poles at $-4, -i, i$. However, I do not think it is kosher to simply state that $\int_{C[0, 3]} \frac{dz}{(z + 4)(z + i)(z - i)} = \int_{C[0, 3]} \frac{dz}{(z + i)(z - i)}.$ If it is, correct me.
Now I evaluate this integral thus
$\int_{C[0, 3]} \frac{dz}{(z + 4)(z + i)(z - i)}$ = $\int_{C[0, 3]} \frac{\frac{1}{z+4}}{(z + i)(z - i)} dz$ = $\int_{C[-i, R_1]} \frac{f_1(z)}{(z + i)} dz + \int_{C[i, R_2]} \frac{f_2(z)}{(z - i)}$. However, I cannot find out the correct values for $f_1, f_2$. Is there a way to do this?
Hint. Use Cauchy's integral formula: for $r>0$ (and $r<2$) $$\int_{C(0, 3)} \frac{dz}{(z + 4)(z + i)(z - i)}=\int_{C(-i, r)} \frac{f_1(z)}{(z + i)} dz + \int_{C(i, r)} \frac{f_2(z)}{(z - i)}=2\pi i f_1(-i)+2\pi i f_2(i)$$ where $f_1(z)=\frac{1}{(z + 4)(z - i)}$ and $f_2(z)=\frac{1}{(z + 4)(z + i)}$.