Determining the general form

For brevity, we define

$$I_n(x) := \int_0^x t^n {\rm e}^t {\rm d}t$$

Integrating by parts,

$$I_n(x) = \int_0^x t^n {\rm d}({\rm e}^t) = x^n {\rm e}^x - n \int_0^x t^{n-1} {\rm e}^t {\rm d}t =: x^n {\rm e}^x - n I_{n-1}(x)$$ (1)

By direct integration,

$$I_0(x) = {\rm e}^x - 1$$

Hence, by (1),

$$I_1(x) = x {\rm e}^x - {\rm e}^x + 1$$

$$I_2(x) = x^2 {\rm e}^x - 2 x {\rm e}^x + 2 {\rm e}^x - 2$$

Looking at the above results, we can make the conjecture

$$I_n(x) = F_n(x) {\rm e}^x - (-1)^n n!$$ (2)

where $F_n(x)$ is a polynomial of degree $n$ in $x$. We can prove our conjecture by induction,

$$I_{n+1}(x) = x^{n+1} {\rm e}^x - (n+1) I_n$$

$$= x^{n+1} {\rm e}^x - (n+1) (F_n(x) {\rm e}^x - (-1)^n n!)$$

$$= (x^{n+1} - (n+1) F_n(x)) {\rm e}^x + (-1)^n (n+1) n!$$

$$= F_{n+1}(x) {\rm e}^x - (-1)^{n+1} (n+1)!$$

$$= I_{n+1}(x)$$

Hence, our conjecture is proved and so $I_n(x)$ has the general form as stated in (2).