Prove that $(n+1)\int_0^1{\frac{x^n}{x+1}dx}\to\frac12$ Let $I_n=\int_0^1{\frac{x^n}{x+1}dx}$ , $n>0$
Show that $\lim_{n->\infty}{(n+1)I_n} = \frac{1}{2}$
All I could do was to show that the $I_n$ is decreasing.

You may just integrate by parts,
$$
\begin{align}
I_n=\int_0^1\frac{x^n}{x+1}dx&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{(x+1)}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\\\
&=\frac1{2(n+1)}+\color{blue}{\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx}
\tag1
\end{align}
$$ then observe that
$$
0\leq \color{blue}{\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx}\leq \frac{1}{(n+1)}\frac{1}{(0+1)^2}\int_0^1x^{n+1}dx=\frac{1}{(n+1)(n+2)}\tag2
$$
Then using $(1)$ and $(2)$ gives easily 

$$ \lim_{n \to +\infty}nI_n=\frac12.$$

