Evaluation of an Integral $x^{x+1}$ Involving Limits I would like to evaluate this weird looking integral a friend showed me:

$\displaystyle \lim_{n \to \infty}  n^2\int_0^{1/n} x^{x+1}\, dx$

I thought that Frullani Integral formula could help, I simply do not know where to begin.
 A: I`ll propose a different approach using functions sequences:
We can substitute $t=xn$,$dt=ndx$ and we have : 
$$\lim_{n\to
\infty}n^2\int_0^{1/n}x^{x+1}dx=\lim_{n\to
\infty}\int_0^{1/n}n^2x^{x+1}dx=\lim_{n\to
\infty}\int_0^{1}n^2(\frac t n)^{t/n+1}dx=$$
$$=\lim_{n\to
\infty}\int_0^{1}n(\frac t n)^{t/n+1}dt=\lim_{n\to
\infty}\int_0^{1}t(\frac t n)^{t/n}dt$$
Looking at the functions sequence $f_n(x)=x(\frac x n)^{x/n} \to x$ pointwise in $[0,1]$.Also, the sequence $f_n$ is monotonically decreasing(with $n$) , continous, and converges pointwise to a continous function in $[0,1]$ so it converges uniformly to $x$,according to Dini`s theorem.
This means, we can move the limit inside the integral:
$$\lim_{n\to
\infty}\int_0^{1}t(\frac t n)^{t/n}dt=\int_0^{1}\lim_{n\to
\infty}t(\frac t n)^{t/n}dt=\int_0^{1}t dt = \frac 1 2$$
A: You can write
$$
x^{x+1} = x e^{x \log (x)}= x + x^2 \log(x) + \frac{x^3}{2} \log^2(x) + \cdots
$$
Integrating will give you a series
$$
I = \int_0^{1/n} x^{x+1} \; dx = \frac{1}{2n^2} + \frac{-1-3 \log(n)}{9n^3} + \frac{1 +4 \log(n) + 8 \log^2(n)}{64n^4} + \cdots
$$
for large $n$ the first term dominates. Then
$$
n^2 I = \frac{1}{2} + \frac{-1-3 \log(n)}{9n} + \frac{1 +4 \log(n) + 8 \log^2(n)}{64n^2} + \cdots
$$
which goes to $1/2$ as $n\to \infty$. Perhaps this is a fluke but it gives more evidence that the answer is $1/2$, which it appears to be from numerical integration and plotting the function.
