If $I_n={\int }_0^1{x}^n{e}^{x}dx$, then $I_n+n{I}_{n-1}$ is equal to:

Solution

1. Apply integration by parts to $I_n={\int }_0^1{x}^n{e}^{x}dx$ with $u=x^n$ and $dv=e^{x}dx$.
2. $I_n=[x^n{e}^x{]}_0^1-{\int }_0^{1}n{x}^{n-1}{e}^{x}dx$.
3. $I_n=e-n{I}_{n-1}$.
4. Rearranging gives $I_n+n{I}_{n-1}=e$.
5. Hence the answer is (A).