= Solution
Differentiate the <velocity potential> in polar coordinates:
$$
\boxed{u_r=U(1-a^2/r^2)\cos\theta,\qquad u_\theta=-U(1+a^2/r^2)\sin\theta+\frac\kappa{2\pi r}}.
$$
At $r=a$ the normal velocity is zero. At large radius these components tend to $U\cos\theta,-U\sin\theta$, the polar components of uniform flow in the positive $x$ direction. The circulation around any concentric circle is $\int_0^{2\pi}u_\theta r\,d\theta=\kappa$. Although the potential's circulation term is multivalued, its velocity is single-valued and irrotational away from the excluded axis.
For steady inviscid constant-density flow, <Bernoulli equation> gives $p+\rho|u|^2/2=p_\infty+\rho U^2/2$. On the cylinder, $u_\theta=-2U\sin\theta+\kappa/(2\pi a)$, so
$$
\boxed{p(a,\theta)=p_\infty+\frac\rho2\left[U^2-\left(-2U\sin\theta+\frac\kappa{2\pi a}\right)^2\right]}.
$$
In the displayed force integral, every constant or $\sin^2\theta$ term has zero cosine and sine first moments. The cross term in pressure is $\rho U\kappa\sin\theta/(\pi a)$. Therefore, using exactly the positive-sign convention printed,
$$
\boxed{F_x=0,\qquad F_y=\rho U\kappa}
$$
per unit axial length, since $\int_0^{2\pi}\sin^2\theta\,d\theta=\pi$.
With the usual normal directed outward from the cylinder into the fluid, pressure exerts traction $-pn$, so the physical force of the fluid on the cylinder is instead $\boxed{(0,-\rho U\kappa)}$. The positive integral printed in the question has the opposite sign. This makes the convention explicit and is consistent with the <Kutta–Joukowski theorem> for positive counterclockwise circulation.
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