Finding $\lim\limits_{n \to \infty }n\int_{0}^{1}\frac{x^{n}}{1+x+x^{n}}dx$ $$\lim_{n \to \infty }n\int_{0}^{1}\frac{x^{n}}{1+x+x^{n}}dx$$
I factorize $x^n$ then I tried with $u=1+x^{1-n}$ but I didn't get too far. Also I tried to get something from $0<x<1$. I know that if $x$ is from $(0,1)$ then $x^n$ tends to $0$ as $n \to \infty$.
 A: In my proof, I shall use Lebesgue's Dominated Convergence Theorem.
We have (I ignore the $dx$ for brevity)
$$
\int_0^1\frac{nx^n}{1+x+x^n}
=\int_0^1\frac{x(nx^{n-1}+1)-x}{1+x+x^n}
=\int_0^1\frac{x(nx^{n-1}+1)}{1+x+x^n}-\int_0^1\frac{x}{1+x+x^n}.
$$
Using IP with $u(x)=x$ and $v'(x)=nx^{n-1}+1$ for the first integral, we get
$$
I_{1,n}=\ln(3)-\int_0^1\ln(1+x+x^n).
$$
Now, since $0\leq\ln(1+x+x^n)\leq x+x^n\leq x+1$ on $(0,1)$ (here I use the inequality $\ln(1+x)\leq x$ for $x>0$)
and $x\mapsto x+1$ is integrable on $(0,1)$, applying DCT yields
$$
\lim_n\int_0^1\ln(1+x+x^n)=\int_0^1\lim_n\ln(1+x+x^n)=\int_0^1\ln(1+x)=\ln(4)-1.
$$
Thus, $I_{1,n}\to\ln(3)-\ln(4)+1$. For the other one, since
$0\leq x/(1+x+x^n)\leq 1$ on $(0,1)$, applying DCT yields,
$$
\lim\int_0^1\frac{x}{1+x+x^n}=\int_0^1\lim\frac{x}{1+x+x^n}=\int_0^1\frac{x}{1+x}=1-\ln(2).
$$
I leave the job of combining these for you :)
A: $$
\begin{align}
\lim_{n\to\infty}n\int_0^1\frac{x^n}{1+x+x^n}\,\mathrm{d}x
&=\lim_{n\to\infty}\int_0^1\frac{x^{1/n}}{1+x^{1/n}+x}\,\mathrm{d}x\tag1\\
&=\int_0^1\frac1{2+x}\,\mathrm{d}x\tag2\\[3pt]
&=\log\left(\frac32\right)\tag3
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x^{1/n}$
$(2)$: Dominated Convergence
$(3)$: evaluate

Heuristics
Why did I use the substitution $x\mapsto x^{1/n}$? The function $nx^n$ has weight $\frac{n}{n+1}$, and all that weight gets concentrated near $1$ as $n\to\infty$. If we didn't have the $x^n$ in the denominator, then we could simply note that $\frac1{1+x}=\frac12$ near $1$ and see that the integral would be $\frac12$. The $x^n$ in the denominator forces us to look closer at what happens near $1$. The map $x\to x^{1/n}$ moves the points of $[0,1]$ closer to $1$; thus, it spreads the action of $\frac{nx^n}{1+x+x^n}$ from near $1$ further down on $[0,1]$. The thing that told me that this was the right thing is that the factor of $n$ disappeared. Once we perform the substitution, its easy to see that for all $n\gt1$,
$$
\frac{x^{1/n}}{1+x^{1/n}+x}\le\frac1{1+x}
$$
which allows us to use Dominated Convergence.

A Far More Basic Approach
$$
\begin{align}
\lim_{n\to\infty}n\int_0^1\frac{x^n}{1+x+x^n}\,\mathrm{d}x
&=\lim_{n\to\infty}\int_0^1\frac{x^{1/n}}{1+x^{1/n}+x}\,\mathrm{d}x\tag4\\
&=\int_0^1\frac1{2+x}\,\mathrm{d}x\tag5\\[3pt]
&=\log\left(\frac32\right)\tag6
\end{align}
$$
Explanation:
$(4)$: substitute $x\mapsto x^{1/n}$
$(5)$: the difference vanishes
$(6)$: evaluate
Discussion of $\mathbf{(5)}$:
$$
\begin{align}
\int_0^1\left[\frac1{2+x}-\frac{x^{1/n}}{1+x^{1/n}+x}\right]\mathrm{d}x
&=\int_0^1\frac{(1+x)\left(1-x^{1/n}\right)}{(2+x)\left(1+x^{1/n}+x\right)}\,\mathrm{d}x\tag7\\
&\le\frac12\int_0^1\left(1-x^{1/n}\right)\mathrm{d}x\tag8\\[6pt]
&=\frac1{2n+2}\tag9
\end{align}
$$
Explanation:
$(7)$: subtract
$(8)$: $\frac{1+x}{(2+x)\left(1+x^{1/n}+x\right)}\le\frac12$
$(9)$: evaluate
