The integral is $\int_{\gamma}\frac{1}{z^{2}-1}dz$ along the path $\gamma(t)=2e^{ti},\;t\in[0,2\pi]$ Which I attempt to do by parts: \begin{equation*} \int_{\gamma}\frac{1}{z^{2}-1}dz= \frac{1}{2}[\int_{\gamma}\frac{1}{z-1}dz-\int_{\gamma}\frac{1}{z+1}dz] \end{equation*} And plugging in $z=\gamma(t)=2e^{it}$ gives \begin{align*} \frac{1}{2}[\int_{\gamma}\frac{1}{z-1}dz-\int_{\gamma}\frac{1}{z+1}dz]&= \frac{1}{2}[\int_{0}^{2\pi}\frac{1}{2e^{it}-1}dz- \int_{0}^{2\pi}\frac{1}{2e^{it}+1}dz] \end{align*} For which we perform the u subsitution \begin{equation*} u_{1}=2e^{ti}-1,\; u_{2}=2e^{ti}+1\Rightarrow dt=\frac{du_{1}}{2ie^{ti}}= \frac{du_{2}}{2ie^{ti}} \end{equation*} for both, yielding \begin{equation*} \frac{1}{2}[\int_{t=0}^{t=2\pi}\frac{1}{u}du- \int_{t=0}^{t=2\pi}\frac{1}{u}du]= \frac{1}{2}\log(u)|_{t=0}^{t=2\pi}-\frac{1}{2}\log(u)|_{t=0}^{t=2\pi} \end{equation*} And since $u_{1}(t=0)=1,\; u_{2}(t=2\pi)=2e^{2\pi i}-1=1$ and $u_{2}(t=0)=3,\; u_{2}(t=2\pi)=2e^{2\pi i}+1=3$ we have \begin{equation*} \frac{1}{2}\log(u)|_{1}^{1}-\frac{1}{2}\log(u)|_{3}^{3}=0 \end{equation*} Also justifying ignoring the potential branch cut problems with log. My issue is that I assume that log will be well defined (i.e. its argument will be in the strip of the complex plane $(-\pi i,\pi i]$. Is this still ok?
Note: I asked about the same integral on a different path here: Difficulty evaluating complex integral and made the judgement that this question because the path is different and I have a specific question about my computation. Thanks!