How do I find $\;\lim_{n\to\infty}n\int_{0}^{1} \frac{x^n}{x^2+1} dx\;$? I'm new here so please, try to bear my poor formatting skills. Here's the problem :

$I_{n}$ is considered a sequence where $n\in N^*$
$$
I_{n} = \int_{0}^{1} \frac{x^n}{x^2+1} dx\ , (\forall)n\in N^*
$$
a) Calculate $I_{2}$
b) Prove that $I_{n+2} + I_{n} = \frac{1}{n+1},(\forall)n\in N^*$
c) Calculate $\lim_{n \to +\infty} nI_{n}$

I have done both a) and b), but I don't exactly know how to solve c).
Any help is appreciated !
 A: $$\color{blue}{I_{n+2} + I_n = \frac 1{n+1}}$$
Let first define $\color{blue}{I_{(n\to\infty)} = I}$
Here, I believe that there shouldn't be any difference for a value $I_n$ and $I_{n+2}$ as $n\to\infty$ they must be same because $n$ is already a very very big number then any small increment like(2) shouldn't make difference provided $n$ is very big number.
$$\begin{align*}
I_{n+2} + I_n
& = \frac 1{n+1}\\
&\lim_{n\to\infty}nI_{n+2} + \lim_{n\to\infty}nI_n = \lim_{n\to\infty}\frac n{n+1}\\
& \text{Now, By intuition we can imagine this that when $n\to\infty$ then $I_{n+2} = I_n$ = I}\\
&\text{(As, from above blue color } I_{(n\to\infty)} = I)\\
&2\times\lim_{n\to\infty}nI = \lim_{n\to\infty}\frac n{n+1} = 1\\
& \implies \lim_{n\to\infty}nI = \frac 12
\end{align*}$$
A: We know $x^n \geq x^{n+2}$ in the range $[0,1]$, so that
$I_n \geq I_{n+2}$, implying
$$I_n \geq \frac{1}{2} \frac{1}{n+1}.$$
Similarly
$$I_n \leq \frac{1}{2} \frac{1}{n-1}.$$
The resulting upper and lower bounds for $n I_n$
each approach 1/2.
