Find $\int x^ne^xdx$ 
Find $\displaystyle \int x^ne^xdx$.

I haven't been able to figure this out. Do I just keep repeatedly applying integration by parts?
 A: May I recommend integration by undetermined coefficients, where given some $n$, we know the solution will take the form $P(x)e^x + C$, where $P(x)$ is some polynomial of degree $n$. (It really can't be anything else). So write $P(x)$ as $a_0 + a_1x + ... + a_nx^n$, take the derivative, and back out the coefficients.
Or if you want to get practice, use integration by parts. But that's really nasty.
A: Finding an explicit formula for the antiderivative will involve the incomplete Gamma function.
What you could do (and what is usually standard) is get a recurrence relation.
Let $$I_n = \int x^n e^x dx, \ n\in \mathbb{N}\cup{0}$$ Integration by parts yields 
$$I_n = x^n e^x -n \int x^{n-1} e^x dx = x^n e^x - n I_{n-1}.$$
The first term in this recurrence relation is $$I_0 = e^x.$$
A: Here is another approach for definite integrals:
Let $I(t) = \int_a^b e^{tx} dx = {1 \over t} (e^{tb}-e^{ta})$.
Then
$I^{(n)}(1) = \int_a^b x^n e^{x} dx$.
A: Notice, integrating by parts as follows  $$\int x^ne^x\ dx$$
$$=x^n\int e^x\ dx-\int nx^{n-1}e^x\ dx$$
$$=x^ne^x-nx^{n-1}e^x+n\int (n-1)x^{n-2}e^x\ dx$$
$$=x^ne^x-nx^{n-1}e^x+n(n-1)x^{n-2}e^x-n(n-1)\int (n-2)x^{n-3}e^x\ dx$$
$$................$$
$=\left(x^n-nx^{n-1}+n(n-1)x^{n-2}-n(n-1)(n-2)x^{n-3}+\ldots +(-1)^{n+1}n!\right)e^x+C$
$\forall \ \ n\in N$
