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

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.

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.

• This is not a duplicate of math.stackexchange.com/questions/21516/…, since in the present question $n$ is an integer. Commented Jun 20, 2011 at 14:36
• The "answer" to the other question involves the gamma function, which I don't think is necessary when n is a positive integer.
– Tim
Commented Jun 20, 2011 at 14:48
• Why do you say it's not a solution? The answer obviously must contain all the powers of $x^k$ for $0 \leq k \leq n$ so the solution can't be much simpler than this. E.g. for $x > 0$ you might try $\left((-\partial_{\alpha})^n \int e^{-\alpha x}\right) \Big\vert_{\alpha = 1}$ but now you have to differentiate $n$ times so this is essentially the same thing. Commented Jun 20, 2011 at 15:06

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 $$\int {x^n e^x dx} = \bigg[\sum\limits_{k = 0}^n {( - 1)^{n - k} \frac{{n!}}{{k!}}x^k } \bigg]e^x + C.$$

• Correct me if I'm wrong but in the general case with $a$ non-zero real or imaginary, this holds: $\int {x^n e^{ax} dx} = \frac{1}{a^{n+1}}\bigg[\sum\limits_{k = 0}^n {( - 1)^{n - k} \frac{{n!}}{{k!}}(ax)^k } \bigg]e^{ax} + C$ ? Is this true for $a$ being any complex number as well? Commented Dec 3, 2022 at 16:59

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, $$\int_0^X x^n e^x\, dx = (-1)^n n! (e^X - 1) + \sum_{j=1}^n (-1)^{n-j} \frac{n!}{j!} X^j e^X$$

• Interesting coincidence! We must have been thinking about this in a similar way as I also posted this answer on Mike Spivey's follow up question. (Although it is admittedly more suited as a response here) See math.stackexchange.com/questions/46733/… Commented Jun 21, 2011 at 19:04

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.

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

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:

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

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

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

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

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

$$k+1=n$$

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