Calculate with Cauchy integral formula Can you please check if i did any mistakes.
a)
$$\int_{\gamma_{B_2(0)}} \frac{z^7+41}{1-z}$$
$\frac{z^7+41}{1-z}$ = $ \frac{-z^7-41}{z-1}$ = $\frac{f(z)}{z-1}$
with the Cauchy integral formula i get $2 \pi i  \cdot (-42) = -84\pi i$
b)
$$\int_{\gamma_{B_r(\frac{i}{2})}} \frac{1}{z^2+1}$$
(1) Let $r = \frac{1}{4}$ then $\frac{1}{z^2+1}$ is holomorphic and there exists a holomorphic antiderivative on $\mathbb C\setminus\{i,-i\}$ so therefore the integral is $0$.
(2)Let $ r = 1$ then we have $ \frac{1}{z^2+1} = \frac{1}{(z+i)(z-i)}=\frac{\frac{1}{z+i}}{z-i} = \frac{f(z)}{z-i}$ . With the CIF i get $2\pi i \cdot f(i) = \pi i$
(3)Let $r=2$ then i get $\pi i$ aswell which is probably wrong 

For $r=2$, both poles are enclosed.  We can write
$$\frac{1}{z^2+1}=\frac{1/2i}{z-i}-\frac{1/2i}{z+i}$$
and apply Cauchy's Integral Formula as 
$$\begin{align}
\oint_{B_{2}(i/2)}\frac{1}{z^2+1}\,dz&=\oint_{B_{2}(i/2)}\left(\frac{1/2i}{z-i}-\frac{1/2i}{z+i}\right)\,dz\\\\
&=\oint_{B_{2}(i/2)}\frac{1/2i}{z-i}\,dz-\oint_{B_{2}(i/2)}\frac{1/2i}{z+i}\,dz\\\\
&=\frac1{2i}-\frac{1}{2i}\\\\
&=0
\end{align}$$
