Finding $\lim_{n \to \infty} \left( 1 + 2\int_0^1 \frac{x^n}{x+1} dx \right)^n$ 
I have to evaluate
  $$\lim_{n \to \infty} \left( 1 + 2\int_0^1 \frac{x^n}{x+1} dx \right)^n. $$

My progress: Since $x \in (0, 1)$ we can use the series expansion of $\frac{1}{1+x} = 1-x+x^2-x^3+...$
Evaluating that integral in the parantheses (which I shall call $I_n$) gives
$$I_n = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{n+k} = (-1)^n(\log{2} - A_n) $$
where $A_n$ is the nth partial sum of the alternating harmonic series, $A_n = \sum_{k=1}^n \frac{(-1)^{k+1}}{k}.$
Since $\log{2} - A_n$ goes to 0, it's enough to compute
 $$2 \lim_{n \to \infty} n(-1)^n(\log{2} - A_n) $$
This is where I got stuck. Any ideas? 

$$|B_n|=|\int_0^1\dfrac{x^n}{1+x} \text{d}x| \leq \int_0^1 x^n\text{d}x=\dfrac{1}{n+1}$$
$$E_n=(1+2B_n)^n=\exp(n\log(1+2B_n))=\exp(n\times(2B_n+O(B_n^2))$$
so $$\lim E_n=\exp(\lim 2nB_n)$$
and $$nB_n=n\int_0^1 \dfrac{x^n}{1+x} \text{d}x=n\int_0^1\dfrac{u}{1+u^{1/n}}u^{(1-n)/n}\times \dfrac{1}{n}\text{d}u=\int_0^1\dfrac{u^{1/n}}{1+u^{1/n}}\text{d}u$$
so by dominated convergence theorem $\lim E_n=\exp(1)$
