Without Cauchy Integral Formula or Residues: $\int_{L}\frac{dz}{z^{2}+1}$ I am currently working on the following:
Prove that $\displaystyle \int_{L}\frac{dz}{z^{2}+1} = 0$ if $L$ is any closed rectifiable simple curve on the outside of the unit disc, i.e., $L$ is contained in the region $|z|>1$. Show that the equality is in general false for arbitrary closed rectifiable simple curves that miss zeros of $z^{2}+1$.
Partial fraction decomposition yields that our integrand is not analytic at $z = \pm i$ (surprise, surprise):
$\displaystyle \int_{L} \frac{dz}{z^{2}+1} = \int_{L} \left( \frac{i}{2(z+i)} - \frac{i}{2(z-i)}\right)dz$.
Now, FYI, I am not allowed to use Cauchy's Integral Formula or Residues to solve this problem. Please do not post answers using either of those two methods, as they are useless to me.
I'm thinking that for the first part, Cauchy's Theorem would work, although Cauchy's Theorem requires the function to be analytic in a simply connected domain. I think that a simply closed rectifiable curve on a region of $\mathbb{C}$ where the function is analytic would form a simply connected domain, but I'm not sure. Update: I am having the hardest time with this part of the problem. I asked my professor, he said the $|z|>1$ allows us to use Cauchy's Theorem for multiple contours. But, I'm not sure how that would work, since we can't draw circles around each of the singularities, because neither of them are in the region. This is the part I would most appreciate an answer for 
For the second part, I'm not entirely sure what is being asked. When they say "miss zeros of $z^{2}+1$", do they mean curves around such zeros? Or do they mean curves that don't go anywhere near them? I'm a bit confused. Update: I now understand this part
Please help.  
 A: Consider the paths shown below

Cauchy's Theorem says that the integral along the paths around the green and yellow regions, following the arrows inside those regions, are both $0$ since
$$
f(z)=\frac1{z^2+1}\tag{1}
$$
is analytic inside those regions.
The integrals along the circular paths around the pink regions, following the arrows inside those regions, are computed explicitly below. The integral along the circular path around the pink region containing $i$ is $\pi$. The integral along the circular path around the pink region containing $-i$ is $-\pi$.
The integral along the path inside the purple region, following the arrows inside that region is the sum of the integrals around the pink, green, and yellow regions since the integrals along all the other contours cancel.
Possibly making the pink circles smaller, if needed, and modifying the contour inside the purple region, we can see that the integral
$$
\oint_\gamma\frac1{z^2+1}\,\mathrm{d}z=\pi-\pi=0\tag{2}
$$
for $\gamma$ contained in the region where $\left|z\right|\gt1$.
For the second part of the question, consider either of the integrals in $(4)$ or $(6)$.

Pink Region Containing $\boldsymbol{i}$
The circular path around the pink region containing $i$ is
$$
\gamma(\theta)=i+re^{i\theta}\tag{3}
$$
where $0\le\theta\lt2\pi$ and $r$ can be arbitrarily small.
$$
\begin{align}
\oint_\gamma\frac1{z^2+1}\,\mathrm{d}z
&=\oint_\gamma\frac i2\left(\frac1{z+i}-\frac1{z-i}\right)\mathrm{d}z\tag{4a}\\
&=\color{#C0C0C0}{\frac i2\oint_\gamma\frac1{z+i}\,\mathrm{d}z}-\frac i2\oint_\gamma\frac1{z-i}\,\mathrm{d}z\tag{4b}\\
&=-\frac i2\int_0^{2\pi}\frac1{re^{i\theta}}\,ire^{i\theta}\,\mathrm{d}\theta\tag{4c}\\
&=\frac12\int_0^{2pi}\mathrm{d}\theta\tag{4d}\\[9pt]
&=\pi\tag{4e}
\end{align}
$$
Explanation:
$\text{(4a)}$: partial fractions
$\text{(4b)}$: $\frac1{z+i}$ is analytic inside this region so Cauchy says the integral is $0$
$\text{(4c)}$: use the parameterization of $\gamma$: $z-i=re^{i\theta}$
$\text{(4d)}$: cancel
$\text{(4e)}$: evaluate

Pink Region Containing $\boldsymbol{-i}$
The circular path around the pink region containing $-i$ is
$$
\gamma(\theta)=-i+re^{i\theta}\tag{5}
$$
where $0\le\theta\lt2\pi$ and $r$ can be arbitrarily small.
$$
\begin{align}
\oint_\gamma\frac1{z^2+1}\,\mathrm{d}z
&=\oint_\gamma\frac i2\left(\frac1{z+i}-\frac1{z-i}\right)\mathrm{d}z\tag{6a}\\
&=\frac i2\oint_\gamma\frac1{z+i}\,\mathrm{d}z\color{#C0C0C0}{-\frac i2\oint_\gamma\frac1{z-i}\,\mathrm{d}z}\tag{6b}\\
&=\frac i2\int_0^{2\pi}\frac1{re^{i\theta}}\,ire^{i\theta}\,\mathrm{d}\theta\tag{6c}\\
&=-\frac12\int_0^{2pi}\mathrm{d}\theta\tag{6d}\\[9pt]
&=-\pi\tag{6e}
\end{align}
$$
Explanation:
$\text{(6a)}$: partial fractions
$\text{(6b)}$: $\frac1{z-i}$ is analytic inside this region so Cauchy says the integral is $0$
$\text{(6c)}$: use the parameterization of $\gamma$: $z+i=re^{i\theta}$
$\text{(6d)}$: cancel
$\text{(6e)}$: evaluate

A: Here is an idea: let $\;L\;$ be any closed, rectifiable and simple curve completely outside the unit circle. Then, both on $\;L\;$ and on the domain it encloses (and this is a simply connected set) the function $\;f(z)=\frac1{z^2+1}\;$ is analytic and it thus has a primitive function (which you can easily call "a potential function" if you think of this as a two variable function). 
Thus, then integral of $\;f\;$ only dependends on its end points and this means $\;\oint_L f(z)\,dz=0\;$ as $\;L\;$ is closed.
The primitive can easily be calculated in this particular case by doing partial fractions:
$$\frac1{z^2+1}=\frac1{(z-i)(z+i)}=\frac1{2i}\left(\frac1{z-i}-\frac1{z+i}\right)\implies$$
$$\int\frac{dz}{z^2+1}=\frac1{2i}\log\frac{z-i}{z+i}+K$$
and the primitive is well defined, analytic on the domain mentioned about.
Observe that no residues or Cauchy Theorem is used, we're only basing our development mainly on our knowledge of real analysis in $\;\Bbb R^2\;$.
