Let function F be defined as F(x) = ∫₁ˣ ettdt, x > 0, then the value of the integral ∫₁ˣ ett+adt, where a > 0, is :

The function F(x) is hard to integrate directly. It is easier to use Leibniz Integral rule instead. Let G(x) = ∫₁ˣ ett+adt. ∴G'(x) = exx+ad(x)dx - e11+ad(1)dx ∴G'(x) = exx+a with a > 1 from the limits. Similarly, F'(x) = exx with x > 1. Also, from F'(x), we can get F'(x+a) = ex+ax+a ⇒ e-aF'(x+a) = exx+a On integrating from 1 to x, we get e-aF(x+a) + c = ∫₁ˣ ett+adt = G(x) To determine the integration constant c, we substitute the boundary condition x=1 which gives e-aF(1+a) + c = ∫₁¹ ett+adt ⇒ e-aF(1+a) + c = 0 ⇒ c = -e-aF(1+a). Therefore, G(x) = e-a(F(x+a) - F(1+a))