Abstract The Volterra integral equation (1) x(t)=a(t)+∫0tB(t,s)x(s)dswith a kernel of the form B(t,s)=p(t)q(s) is investigated, where a, p, and q are functions that are defined a.e. on an interval [0,T] and are measurable. The main result of this paper states that if qa is Lebesgue integrable on [0,T], the sign of B(t,s) does not change for almost all (t,s), and if there is a function f that is continuous on [0,T], except possibly at countably many points, with B(t,t)=f(t) a.e. on [0,T], then the function x defined by (2) x(t)≔a(t)+∫0tR(t,s)a(s)ds,where (3) R(t,s)≔B(t,s)e∫stB(u,u)du,solves (1) a.e. on [0,T]. Three diverse examples illustrate the efficacy of using (2) and (3) to calculate solutions of (1). Two of the examples involve singular kernels: the solution of one of them is nowhere continuous on (0,T).