The first part of the arc [tex]C[/tex] can be parameterized by
[tex]C_1:\mathbf r(t)=\langle 2\cos t,2\sin t\rangle[/tex]
with [tex]0\le t\le\dfrac\pi2[/tex], and the second part by
[tex]C_2:\mathbf s(t)=(1-t)\langle0,2\rangle+t\langle4,3\rangle=\langle4t,2+t\rangle[/tex]
with [tex]0\le t\le 1[/tex].
Now the line integral can be computed.
[tex]\displaystyle\int_{C=C_1\cup C_2}(x^2\,\mathrm dx+y^2\,\mathrm dx)=\int_0^{\pi/2}\left((2\cos t)^2\dfrac{\mathrm d}{\mathrm dt}(2\cos t)+(2\sin t)^2\dfrac{\mathrm d}{\mathrm dt}(2\sin t)\right)\,\mathrm dt+\int_0^1\left((4t)^2\dfrac{\mathrm d}{\mathrm dt}(4t)+(2+t)^2\dfrac{\mathrm d}{\mathrm dt}(2+t)\right)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{\pi/2}(-8\cos^2t\sin t+8\sin^2t\cos t)\,\mathrm dt+\int_0^1(64t^2+(2+t)^2)\,\mathrm dt[/tex]
[tex]=\displaystyle-8\int_0^{\pi/2}\sin t\cos t(\cos t-\sin t)\,\mathrm dt+\int_0^1(65t^2+4t+4)\,\mathrm dt[/tex]
[tex]=\displaystyle-4\int_0^{\pi/2}\sin2t(\cos t-\sin t)\,\mathrm dt+\int_0^1(65t^2+4t+4)\,\mathrm dt[/tex]
[tex]=\displaystyle-2\int_0^{\pi/2}(\sin t+\sin3t-\cos t+\cos3t)\,\mathrm dt+\int_0^1(65t^2+4t+4)\,\mathrm dt[/tex]
[tex]=\displaystyle-2\left(-\cos t-\frac{\cos3t}3-\sin t+\frac{\sin3t}3\right)\bigg|_{t=0}^{t=\pi/2}+\left(\frac{65}3t^3+2t^2+4t\right)\bigg|_{t=0}^{t=1}[/tex]
[tex]=0+\dfrac{83}3[/tex]
[tex]=\dfrac{83}3[/tex]
