Evaluate the contour integral (Most likely without Green's Theorem) $\int_{c}\frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy$ where $C$ is the triangle with vertices at $(5,5), (-5,5),$ and $(0,-5)$ traversed counterclockwise. (Hint: Be careful about the hypotheses of any theorem you use).
My attempt: It appears clear that we cannot use Green's theorem here, as the functions are not defined at the origin. Hence, we can break the line integral into three distinct line integrals. For the line between $(5,5)$ and $(-5,5)$, we can substitute $y=5$ into the integral and solve, for the line between $(0,-5)$ and $(5,5)$ we can substitute $y= 2x-5$, and for the final line we can substitute $y = -2x-5$.
My trouble comes with attempting to evaluate the first integral
$\int_{-5}^5\frac{-5}{x^2+25}dx + \frac{x}{x^2+25}dy$.
Can I simplify this to
$\int_{-5}^5\frac{1}{x+5}dx$?
This would give an integral that does not converge, I believe.
Any help greatly appreciated!

Parameterizing by using the $x$ coordinate as the parameter, we have
$$\begin{align}
\oint_{\partial \mathbb{D}} \left(\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)&=\int_0^5\left(\frac{-(2x-5)}{x^2+(2x-5)^2}\,dx+\frac{x}{x^2+(2x-5)^2}\,2\,dx\right)\\\\&+\int_{5}^{-5}\left(\frac{-5}{x^2+25}\,dx\right)\\\\&+\int_{-5}^0 \left(\frac{-(-2x-5)}{x^2+(2x+5)^2}\,dx+\frac{x}{x^2+(2x+5)^2}\,(-2)\,dx\right)\\\\&=5\int_0^5 \frac{1}{x^2+(2x-5)^2}\,dx\\\\
&+10\int_0^5 \frac{1}{x^2+25}\,dx\\\\
&+5\int_{-5}^0 \frac{1}{x^2+(2x+5)^2}\,dx\\\\
&=10\int_0^5 \left(\frac{1}{x^2+(2x-5)^2}+\frac{1}{x^2+25}\right)\,dx \\\\
&=10\left(\frac{3\pi}{20}+\frac{\pi}{20}\right)\\\\
&=2\pi
\end{align}$$

NOTE: 
We could have exploited Green's Theorem by deforming the original contour with the classical "keyhole" that removes the origin with a circular contour of radius $\epsilon$.  Then, transforming to polar coordinates reveals
$$\begin{align}
\oint_{\partial \mathbb{D}} \left(\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)&=\int_{\sqrt{x^2+y^2}=\epsilon} \left(\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)\\\\
&=\int_0^{2\pi }\frac{\epsilon}{\epsilon^2}\,\epsilon\,d\phi\\\\
&=2\pi
\end{align}$$
as expected!
