Derive the given formula.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^n{e}^{x}dx=x^n{e}^x-n{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{n-1}{e}^{x}dx$

We integrate by parts. Let $u=x^n$ and $v^{\text{'}}=e^x,$ so $u^{\text{'}}=n{x}^{n-1}$ and $v=e^x.$ Then ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^n{e}^{x}dx=x^n{e}^x-n{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{n-1}{e}^{x}dx$

We use the substitution $w=x^n,$ so $dw=n{x}^{n-1}dx.$ Then, ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^n{e}^{x}dx=x^n{e}^x-n{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{n-1}{e}^{x}dx$

We use the substitution $w=x^n{e}^x,$ so $dw=x^n{e}^{x}dx.$ Then, ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^n{e}^{x}dx=x^n{e}^x-n{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{n-1}{e}^{x}dx$

We integrate by parts. Let $u=x^{n-1}$ and $v^{\text{'}}=e^x,$ so $u^{\text{'}}=x^n$ and $v=e^x.$ Then ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^n{e}^{x}dx=x^n{e}^x-n{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{n-1}{e}^{x}dx$

We integrate by parts. Let $u^{\text{'}}=x^n$ and $v=e^x,$ so $u=n{x}^{n-1}$ and $v^{\text{'}}=e^x.$ Then ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^n{e}^{x}dx=x^n{e}^x-n{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{n-1}{e}^{x}dx$