Here's another idea.
Let $P_n(x)$ be the polynomial defined by
$$P_n(x) = \frac{x^n – (1-x)^n}{2x-1},$$
and note that $P_n(x)=(1-x)P_{n-1}(x) + x^{n-1} \qquad (1), $ where $P_1(x)=1.$
Define $I_n = \int_0^{1} P_n(x) dx.$ Integrating (1) between 0 and 1 we obtain
$$I_n = \int_0^{1} (1-x)P_{n-1}(x) dx + \frac{1}{n},$$
and using the symmetry of $P_n(x)$ about 1/2 we note that
$\int_0^{1} (1-x)P_{n-1}(x) dx = \frac{1}{2}I_{n-1}.$ Hence we obtain the recurrence relation for $n > 1$
$$I_n = \frac{1}{2}I_{n-1} + \frac{1}{n}, \textrm{ where } I_1=1. \qquad (2)$$
From which it follows that if $\lim_{n \rightarrow \infty} n I_n$ exists then it must equal 2.
To show that the limit exists we can solve (2) to obtain
$$I_n = \sum_{k=1}^n \frac {1}{k2^{n-k}},$$
and straightforward algebra shows that $(n-1)I_{n-1} – nI_n > 0$ for sufficiently large $n$ (in fact, $n>6$), and since $nI_n$ is bounded below by 0 the result follows.
Just for completeness:
$$(n-1)I_{n-1} – nI_n = \left\lbrace \sum_{k=1}^{n-2} \frac{k}{2^k(n-k)(n-k-1)} \right\rbrace - \frac{n}{2^{n-1}} \quad \textrm{ for } n>2.$$
Comparing the first term in the summation with the negative term we have
$$ \frac{1}{2(n-1)(n-2)} > \frac{n}{2^{n-1}} \quad \textrm{ for } n \ge 13.$$
Hence $\lbrace nI_n \rbrace$ eventually forms a strictly decreasing sequence.