Finding $F(x)$ from $F(kx),$ where $F(x)$ is the antiderivative of the function $f(x)$. I have that $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1$, and I would like to find $F(x)$.
Attempt
Since $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1,$ $F(t) = \alpha_{1}t^{\beta_{1}} + \alpha_{2}t^{\beta_{2}} + \alpha_{3}t^{\beta_{3}} + \alpha_{4}t^{\beta_{4}}.$
Let $t = e^{x}x,$ which means that $F(t) = \alpha_{1}(e^{x}x)^{\beta_{1}} + \alpha_{2}(e^{x}x)^{\beta_{2}} + \alpha_{3}(e^{x}x)^{\beta_{3}} + \alpha_{4}(e^{x}x)^{\beta_{4}} = e^{x}x^{2} - e^{x}x + e^{x} - 1.$
Therefore, $\alpha_{2} = -1, \alpha_{4} = -1, \beta_{2} = 1,$ and $\beta_{4} = 0.$
$F(t) = \alpha_{1}t^{\beta_{1}} - t + \alpha_{3}t^{\beta_{3}} - 1 = e^{x}x^{2} - e^{x}x + e^{x} - 1.$

Ok, let's summarize the comments in an answer. Differentiating expression for $F(xe^x)$ we get $$(e^x+xe^x)f(xe^x) = x(e^x + xe^x)$$ and thus, for $x\neq -1$ we have $$f(xe^x) = x \implies f(x) = W(x)$$ and consequently, $$F(x) = \displaystyle\int_a^x W(t)\ dt$$ Now, to determine $a$: 
$$\begin{align*}
F(xe^x) &= \displaystyle\int_a^{xe^x} W(t)\ dt \\
&= \left[ t = ue^u,\ dt = (u+1)e^u \right] \\
&= \displaystyle\int_{W(a)}^x u(u+1)e^u\ du\\
&= (x^2 - x + 1)e^x - (W(a)^2 - W(a) + 1)e^{W(a)} \implies W(a) = 0 \implies a = 0
\end{align*}$$ so, $$F(x) = \displaystyle\int_0^x W(t)\ dt$$
