How to integrate $ \int x^n e^x dx$?

Question

How can I solve this indefinite integral for an arbitrary integer $n>0$?

$$ \int{x^n e^x dx}$$

I could partially integrate it for small $n$, but that's not really a solution.

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Edit: (TB) This question is closely related to: Is there a closed form solution for $\int x^n e^{cx}$?, but it is more elementary, because $n$ is an integer here.

Answer 1

Hint: Use integration by parts.

EDIT: Try several values of $n$. $$ \int {x e^x dx} = (x - 1)e^x + C $$ $$ \int {x^2 e^x dx} = (x^2 - 2x + 2)e^x + C. $$ $$ \int {x^3 e^x dx} = (x^3 - 3x^2 + 6x - 6)e^x + C. $$ $$ \int {x^4 e^x dx} = (x^4 - 4x^3 + 12x^2 - 24x + 24)e^x + C. $$ $$ \int {x^5 e^x dx} = (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120)e^x + C. $$ Conclude that $$ \int {x^n e^x dx} = \bigg[\sum\limits_{k = 0}^n {( - 1)^{n - k} \frac{{n!}}{{k!}}x^k } \bigg]e^x + C. $$

Answer 2

You could use the generating function approach.
$$ \eqalign{\int_0^X e^{tx} e^x \ dx &= \frac{e^{(1+t)X} - 1}{1+t}\cr &= \sum_{k=0}^\infty (-1)^k t^k \left(e^X -1 + \sum_{j=1}^\infty e^X \,\frac{X^j}{j!} t^j\right)\cr &= \sum_{n=1}^\infty \left((-1)^n (e^X - 1) + \sum_{j=1}^n (-1)^{n-j} \frac{X^j}{j!} e^X \right) t^n\cr}$$ But also $$ \int_0^X e^{tx} e^x \, dx = \sum_{n=0}^\infty \frac{t^n}{n!} \int_0^X x^n e^x \, dx$$ Equating coefficients of $t^n$ from both sides, $$ \int_0^X x^n e^x\, dx = (-1)^n n! (e^X - 1) + \sum_{j=1}^n (-1)^{n-j} \frac{n!}{j!} X^j e^X $$

Answer 3

I find it a little difficult for me to guess the solution by trying several $n$. I would like to do it as following:

$$\begin{align}\int x^ne^xdx&=x^ne^x+(-1)n\int x^{n-1}e^xdx,\qquad n\geq 1\\ \int x^0e^xdx&=e^x\end{align}$$

Then you get the recurrence relation:

$$\begin{align}a_n(x)&=x^ne^x+(-1)na_{n-1}(x),\qquad n\geq 1\\ a_0(x)&=e^x\end{align}$$

With the recursive formula, it may be easier to find the pattern of the result.

Answer 4

$ \int{x^ne^xdx}=e^x\sum^n_{k=0}{{\left(-1\right)}^{n-k}\frac{n!}{k!}x^k}+C$

Solution:

$ \int{xe^xdx}=\ ?$

$ u=x\to \ \frac{du}{dx}=1\to du=dx$

$ dv=e^xdx\to \frac{dv}{dx}=e^x\to \ \int{\frac{dv}{dx}dx}=\int{e^xdx}+C{{\stackrel{C=0}{\longrightarrow}}}v=e^x$

$ \frac{duv}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}\to \ \int{\frac{duv}{dx}dx}=\int{v\frac{du}{dx}dx}+\int{u\frac{dv}{dx}dx}+C_a\to \int{udv}=uv-\int{vdu}+C_b$

$ \int{xe^xdx}=xe^x-\int{e^xdx}+C_1=xe^x-e^x+C=e^x\sum^1_{k=0}{{\left(-1\right)}^{1-k}\frac{1!}{k!}x^k}+C$

$ \int{x^2e^xdx}=?$

$ u=x^2\to du=2xdx$

$ dv=e^xdx{{\stackrel{C=0}{\longrightarrow}}}v=e^x$

$ \int{x^2e^xdx}=x^2e^x-2\int{{xe}^xdx}+C=x^2e^x-2\left(xe^x-e^x+C\right)=x^2e^x-2xe^x+2e^x+C_1=e^x\sum^2_{k=0}{{\left(-1\right)}^{2-k}\frac{2!}{k!}x^k}+C_1$

Now we suppose:

$ \int{x^ke^xdx}=e^x\sum^k_{i=0}{{\left(-1\right)}^{k-i}\frac{k!}{i!}x^k+C}$

$ \int{x^{k+1}e^xdx=\ ?}$

$ u=x^{k+1}\to du=\left(k+1\right)x^kdx$

$ dv=e^xdx{{\stackrel{C=0}{\longrightarrow}v=e^x}}$

$ \int{x^{k+1}e^xdx=\ x^{k+1}e^x-\int{\left(k+1\right)x^ke^xdx}}+C=x^{k+1}e^x-\left(k+1\right)\int{x^ke^xdx}+C=x^{k+1}e^x-\left(k+1\right)e^x\sum^k_{i=0}{{\left(-1\right)}^{k-i}\frac{k!}{i!}x^k+C}=e^x\sum^{k+1}_{i=0}{{\left(-1\right)}^{k-i+1}\frac{(k+1)!}{i!}x^{k+1}+C}$

$ k+1=n$

$ \int{x^ne^xdx}=e^x\sum^n_{i=0}{{\left(-1\right)}^{n-i}\frac{n!}{i!}x^n+C}$