Evaluate $\int _{D}\frac{\sin z}{z^2-z}dz$ 
Evaluate $$\int_{D}\frac{\sin z}{z^2-z}dz$$ where $D$ is $|z-1|=2$

$$\int_{D}\frac{\sin z}{z^2-z}dz=\int_{\gamma_1}\frac{\frac{\sin z}{z-1}}{z}dz+\int_{\gamma_2}\frac{\frac{\sin z}{z}}{z-1}dz$$ where $\gamma_1$ is a closed ball around $0$ and $\gamma_2$ is a closed ball around $1$
Using cauchy integral theorem we get $$\int_{\gamma_1}\frac{\frac{\sin z}{z-1}}{z}dz+\int_{\gamma_2}\frac{\frac{\sin z}{z}}{z-1}dz=2\pi i\cdot 0+2\pi i\cdot \sin(1)$$
Can I use practical fractions and cauchy integral theorem to solve it too?
 A: Partial fraction expansion gives $$\frac{1}{z^2-z}=\frac{1}{z-1}-\frac{1}{z}$$
Then, we can write
$$\oint_{|z-1|=2}\frac{\sin(z)}{z^2-z}\,dz=\oint_{|z-1|=2}\frac{\sin(z)}{z-1}\,dz-\oint_{|z-1|=2}\frac{\sin(z)}{z}\,dz \tag1$$
Note that the integrand of the second integral on the right-hand side of $(1)$ is analytic.  Therefore, Cauchy's Integral Theorem guarantees that $\oint_{|z-1|=2}\frac{\sin(z)}{z}\,dz=0$.  
Using Cauchy's Integral Formula to evaluate the first integral on the right-hand side of $(1)$ reveals $\oint_{|z-1|=2}\frac{\sin(z)}{z-1}\,dz=2\pi i \sin(1)$.
Putting it together yields
$$\oint_{|z-1|=2}\frac{\sin(z)}{z^2-z}\,dz=2\pi i \sin(1)$$
A: Of course. Just write
$$\oint_D \frac{\sin z}{z(z-1)}\mathrm d z=\oint_D \sin z\left (\frac{1}{z-1}-\frac{1}{z} \right)\mathrm d z=\oint_D\frac{\sin z}{z-1}\mathrm d z-\oint_D \frac{\sin z}{z}\mathrm d z$$
Since the latter function is holomorphic on $\mathbb C$ if extended by continuity in $z=0$ (remember that $\lim_{z\to 0}\frac {\sin z}{z}=1$), its integral is zero.  
For the former integral, observe that because of Cauchy's formula,
$$\oint_D\frac{\sin z}{z-1}\mathrm d z=2\pi i\sin (1)$$
