New idea to solve $\int x^n e^x dx$

Question

I have this problem $$\int x^n e^x dx= x^ne^x -nx^{n-1}e^x +n(n-1)x^{n-2}e^x- \cdots+(-1)^nn!e^x $$ my try was to use integration by part . $$I_{n}=\int x^n e^x \, dx=e^x x^n -\int (nx^{n-1})e^x \, dx \\ \to I_n=e^xx^n-nI_{n-1} \\ \to I_n=e^xx^n -n(x^{n-1}e^x- (n-1)I_{n-2})\\ \cdots $$this way get the answer easily , But I am looking for new idea
somewhere I see a matrix form to get the answer ! but I don't get it ...
any help will be appreciated

Answer 1

Define a function

$$F(a)=\int e^{ax}dx=\frac1a e^{ax}$$

Then, the $n$'th derivative $F^{(n)}(a)$ is given by

$$\begin{align} F^{(n)}(a)&=\int x^ne^{ax}dx\\\\ &=\frac{d^n}{da^n}\left(\frac1a e^{ax}\right)\\\\ &=\sum_{k=0}^{n}\binom{n}{k}\frac{d^k}{da^k}\left(\frac1a\right)\frac{d^{n-k}}{da^{n-k}}\left(e^{ax}\right)\\\\ &=\sum_{k=0}^{n}\binom{n}{k}\left((-1)^kk!a^{-k-1}\right)\left(x^{n-k}e^{ax}\right) \end{align}$$

Setting $a=1$ yields

$$\bbox[5px,border:2px solid #C0A000]{\int x^ne^{x}dx=e^x\left(\sum_{k=0}^{n}\binom{n}{k}(-1)^kk!x^{n-k}\right)+C}$$