Integral of the product $x^n e^x$ I would be very pleased if you could give me your opinion about this way of integrating the following expression. I think that it has no issues, but just wanted to confirm:
$$ \int x^n e^x dx $$
$$ e^x = u $$
$$ e^x dx = du $$
$$ \int x^n e^x dx = \int \ln(u)^n du = nu(\ln(u) - 1) + C $$
$$ \int x^n e^xdx = ne^x (x-1) + C $$
Many thanks!

Another variation is using the Differential operator $D_x$ and it's inverse $D_x^{-1}$ denoting indefinite integration.

The  following is valid
  \begin{align*}
D_x^{-1}\left(e^xf(x)\right)=e^x\frac{1}{1+D_x}f(x) \tag{1}
\end{align*}
We obtain with (1)
  \begin{align*}
\int e^x x^n\, dx&=\frac{1}{D_x}\left(e^x x^n\right)\\
&=e^x\frac{1}{1+D_x}x^n\\
&=e^x\sum_{k=0}^\infty(-1)^kD_x^k x^n\tag{2}\\
&=e^x\sum_{k=0}^n(-1)^k\frac{n!}{k!}x^{n-k}
\end{align*}

Comment


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*In (2) we use the  geometric series expansion.

