Integration without using Integration by Parts There was a post asking for help defining $\int x^{2} e^{x} \, \mathit{dx}$ using Integration by Parts.  (See question 2123271.)  A comment was made that the integral could be defined without using Integration by Parts.  "If $f(t) = \int e^{tx} \, \mathit{dx}$, $f^{\prime\prime}(1) = \int x^{2} e^{x} \, \mathit{dx}$."
What is the definition of $f^{\prime}(t)$ and $f^{\prime\prime}(t)$?
The comment on the post states that the value of $f^{\prime\prime}(t)$ at $1$ helps to show that
\begin{equation*}
\int x^{2} e^{x} \, \mathit{dx} = (x^{2} - 2x + 2)e^{x} .
\end{equation*}
Please explain.
 A: The trick is that we know how to compute this integral exactly: 
$$
\int e^{tx}dx
$$ The result is a function of $t$ (including the constant of integration)
$$
\int e^{tx} dx = \frac{1}{t}e^{tx} + C_1(t)
$$  So, this means (assuming everything is mathematically "kosher", which here it is), 
$$
\frac{d}{dt}\int e^{tx}dx = \frac{d}{dt} \left(\frac{1}{t}e^{tx} + C\right) = -\frac{1}{t^2}e^{tx} + \frac{x}{t}e^{tx} + C_2(t)
$$  But on the other hand, (again, assuming we can "pull the derivative inside the integral", which we can here): 
$$
\frac{d}{dt}\int e^{tx}dx = \int\frac{\partial}{\partial t}e^{tx} dx = \int xe^{tx}dx
$$  So we have arrived at the nice formula 
$$
\int xe^{tx} dx = \left(\frac{x}{t} - \frac{1}{t^2}\right)e^{tx} + C_2(t)
$$  You can evaluate this at any value of $t$; $t=1$ will give $\int xe^xdx$.  If you do this trick again (take a second derivative with respect to $t$) and evaluate the result at $t=1$, you can arrive at the formula for $\int x^2e^{x}dx$.  Note that the "constant of integration" $C_1(t)$ can be chosen to be any function of $t$, so after evaluating at some $t$ value, we still have an "arbitrary constant" - if you were trying to evaluate a definite integral by this method, you could choose $C_1(t)\equiv 0$.
A: Let $f(t)=\int e^{tx}\,dx=\frac{e^{tx}}{t}+C$.  Taking the derivative, $f'(t)$, of $f(t)$ with respect to $t$ reveals
$$f'(t)=\int xe^{tx}\,dx=x\frac{e^{tx}}{t}-\frac{e^{tx}}{t^2}+C \tag 1$$
Note that at $t=1$, $(1)$ becomes $f'(1)=\int xe^x\,dx=(x-1)e^x+C$.
Differentiating $(1)$ with respect to $t$ yields
$$f''(t)=\int x^2e^{tx}\,dx=x^2\frac{e^{tx}}{t}-2x\frac{e^{tx}}{t^2}+2\frac{e^{tx}}{t^3}+C\tag 2$$
Evaluating $(2)$ at $t=1$, we obtain
$$f''(1)=\int x^2 e^{x}\,dx=(x^2-2x+2)e^x+C$$
as was to be shown!
