The path [tex]C_2[/tex] is parameterized by
[tex]\mathbf r(t)=(x(t),y(t))=\dfrac{(\cos t,\sin t)}{1+e^t}[/tex]
so that substituting [tex]x(t),y(t)[/tex] into [tex]\mathbf F[/tex] gives
[tex]\mathbf F(x(t),y(t))=\dfrac{(\cos t,\sin t)}{(1+e^t)\sqrt{1-\frac1{(1-e^t)^2}}}[/tex]
Compute the differential of [tex]\mathbf r(t)[/tex]:
[tex]\mathrm d\mathbf r=\dfrac{(-\sin t-e^t(\cos t+\sin t),\cos t+e^t(\cos t-\sin t))}{(1+e^t)^2}\,\mathrm dt[/tex]
The line integral then reduces to
[tex]\displaystyle\int_{C_2}\mathbf F\cdot\mathrm d\mathbf r=\int_0^\infty-\frac{e^t}{(1+e^t)^3\sqrt{1-\frac1{(1+e^t)^2}}}\,\mathrm dt=-\int_0^\infty\frac{e^t}{(1+e^t)^2\sqrt{(1+e^t)^2-1}}\,\mathrm dt[/tex]
To compute the integral, substitute [tex]s=1+e^t[/tex] and [tex]\mathrm ds=e^t\,\mathrm dt[/tex], so the integration domain changes to the interval [tex][2,\infty)[/tex].
[tex]\displaystyle\int_{C_2}\mathbf F\cdot\mathrm d\mathbf r=-\int_2^\infty\frac{\mathrm ds}{s^2\sqrt{s^2-1}}[/tex]
Then substitute [tex]s=\sec r[/tex] and [tex]\mathrm ds=\sec r\tan r\,\mathrm dr[/tex]. Note that in order for this substitution to be reversible, we restrict [tex]r[/tex] to be between -π/2 and π/2, so that the integration domain changes to [tex]\left[\frac\pi3,\frac\pi2\right)[/tex].
[tex]\displaystyle\int_{C_2}\mathbf F\cdot\mathrm d\mathbf r=-\int_{\pi/3}^{\pi/3}\frac{\sec r\tan r}{\sec^2r\sqrt{\sec^2r-1}}\,\mathrm dr=-\int_{\pi/3}^{\pi2}\cos r\,\mathrm dr=\boxed{-1+\frac{\sqrt3}2}[/tex]
