<s>[INST] Evaluate $\displaystyle \int_C \frac{1}{4z^2+4z-3}$ where $C=C^{+}_{1}(0)$. 
Evaluate $\displaystyle \int_C \frac{1}{4z^2+4z-3}$ where $C=C^{+}_{1}(0)$.

I am not sure whether I understand how to use Cauchy's theorem which states that: 
If $f$ is a analytic in a simply connected domain $D$, and $C$ is a simple closed contour lying in $D$, then $\displaystyle \int_C f(z)dz=0$. (From this also follows the deformation of contours).
Anyways, what I have done so far is in decomposing $f(z)=\frac{1}{4z^2+4z-3}=\frac{1}{4}\left(\frac{1}{2z-1}-\frac{1}{2z+3}\right)$.
Now, I look at the singularities of $f$ which are $z=\frac{1}{2},-\frac{3}{2}$. My contour is the circle (with positive orientation) centered at $0$ with radius $1$ (so the integral part of $\frac{1}{4}\frac{1}{2z+3}$ just goes to $0$ by Cauchy). I then make a contour $C_{1/2}$ around the singularity $\frac{1}{2}$ so I can apply the deformation theorem. I'm then left with $\frac{1}{4}{2\pi i}=\frac{\pi}{2}i$. But apparently the answer is $\frac{\pi}{4}i$. Where did I go wrong here? 
Thanks.
 [/INST] $\displaystyle\frac{1}{4}\frac{1}{2z-1}=\frac{1}{8}\frac{1}{z-\frac{1}{2}}$, so by Cauchy Integral Formula for discs,
\begin{equation*}
\frac{1}{8}\int_{B_1(0)}\frac{dz}{z-\frac{1}{2}}=\frac{1}{8}2\pi i=\frac{1}{4}\pi i
\end{equation*}
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