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? 

Substituting $x\to t^{\frac{1}{n}}$ and  using $\lim_{n\to \infty } \, t^{1/n}=1$ we have
$$I_{n}=\int_0^1 \frac{x^n}{1+x}\,dx = \frac{1}{n}\int_0^1 \frac{t^{\frac{1}{n}}}{1+t^{\frac{1}{n}}}\,dt\to \frac{1}{n}\int_0^1 \frac{1}{1+1}\,dt=\frac{1}{2n}$$
Hence
$$(1+2 I_n)^n \to (1+ \frac{1}{n})^n\to e$$
The limit is e, Euler's constant.
