We compute the integral over the smooth parts, \(C_1\) and \(C_2\text{,}\) of \(C\) separately. We first look at \(C_1\text{.}\) The function \(\vect\gamma(t):=(t,t)\text{,}\) \(t\in[-1,1]\) is a parametrisation consistent with the orientation of \(C_1\text{.}\) Hence \(\vect\gamma'(t)=(1,1)\) and therefore
\begin{align*}
\int_{C_1}2xy\,dx+x^2\,dy
\amp=\int_{-1}^12\gamma_1(t)\gamma_2(t)\gamma_1'(t)
+(\gamma_1(t))^2\gamma_2'(t)\,dt\\
\amp=\int_{-1}^12t^2+t^2\,dt\\
\amp=\int_{-1}^13t^2\,dt
=[t^3]_{-1}^1=1-(-1)=2\text{.}
\end{align*}
Next we compute the integral over \(C_2\text{.}\) A regular parametrisation, consistent with the orientation, is given by \(\vect\gamma(t):=\sqrt{2}(\cos t,\sin t)\text{,}\) \(t\in[\pi/4,5\pi/4]\text{.}\) We have \(\vect\gamma'(t)=\sqrt{2}(-\sin t,\cos t)\text{,}\) and so
\begin{align*}
\int_{C_2}2xy\,dx+x^2\,dy
\amp=\int_{\pi/4}^{5\pi/4}2\gamma_1(t)\gamma_2(t)\gamma_1'(t)
+(\gamma_1(t))^2\gamma_2'(t)\,dt\\
\amp=\int_{\pi/4}^{5\pi/4}2(\sqrt{2}\cos t)(\sqrt{2}\sin t)
(-\sqrt{2}\sin t)+2\cos^2 t\sqrt{2}\cos t\,dt\\
\amp=\int_{\pi/4}^{5\pi/4}-4\sqrt{2}\sin^2 t\cos t
+2\sqrt{2}\cos t(1-\sin^2t)\,dt\\
\amp=\int_{\pi/4}^{5\pi/4}-6\sqrt{2}\sin^2 t\cos t
+2\sqrt{2}\cos t\,dt\\
\amp=\bigl[-2\sqrt{2}\sin^3 t+2\sqrt{2}\sin t\bigr]_{\pi/4}^{5\pi/4}
=-2\text{.}
\end{align*}
Hence we have
\begin{align*}
\int_C 2xy\,dx+x^2\,dy
\amp=\int_{C_1}2xy\,dx+x^2\,dy+\int_{C_2}2xy\,dx+x^2\,dy\\
\amp=2-2=0\text{.}
\end{align*}
