<s>[INST] If $I_n=\int_0 ^1{\frac{x^{n+1}}{x+3}}dx$, prove $\lim_{n \to \infty} n I_n=\frac{1}{4}$ 
Given $I_n=\displaystyle\int_0 ^1{\frac{x^{n+1}}{x+3}}\,dx$ for $n\in\mathbb N$, prove that $$\lim_{n \to \infty} nI_n=\frac{1}{4}$$ 

Here is what the limit looks like:
$$\lim_{n \to \infty} n\int_0 ^1{\frac{x^{n+1}}{x+3}}\,dx$$
I am not sure how to go about this exercise, but I tried solving the integral with no success. I integrated by parts once choosing $u = x^{n+1}$ and $v = \frac{1}{x+3}$ and this is what I got.
$$I_n=\ln 4-(n+1)\int_0^1{x^n\ln(x+3)}\,dx$$
I don't know how to proceed. I was asked previously to prove that $I_{n+1}+3I_n=\frac{1}{n+2}$, do you think this can be used somehow? I tried solving for $I_n$ but I don't see how it can help because of the $I_{n+1}$ that is left over? Help me out, please!
 [/INST] By writing $n I_n = \frac{n}{n+2} \cdot (n+2)I_n$ and noting that $\frac{n}{n+2} \to 1$, it suffices to show that $(n+2)I_n$ converges to $\frac{1}{4}$. However, by integration by parts,
\begin{align*}
(n+2) I_n
&= \int_{0}^{1} (n+2)x^{n+1} \cdot \frac{1}{x+3} \, \mathrm{d}x \\
&= \underbrace{\left[ x^{n+2} \cdot \frac{1}{x+3} \right]_{0}^{1}}_{=\frac{1}{4}} + \int_{0}^{1} \frac{x^{n+2}}{(x+3)^2} \, \mathrm{d}x.
\end{align*}
Moreover, from the inequality $0\leq \frac{1}{(x+3)^2} \leq \frac{1}{9}$ which holds true for $x \in [0, 1]$, we get
$$ 0 \leq \int_{0}^{1} \frac{x^{n+2}}{(x+3)^2} \, \mathrm{d}x \leq \int_{0}^{1} \frac{x^{n+2}}{9} \, \mathrm{d}x = \frac{1}{9(n+3)}. $$
Combining altogether, we get
$$ \frac{1}{4} \leq (n+2)I_n \leq \frac{1}{4} + \frac{1}{9(n+3)} $$
and therefore $(n+2) I_n \to \frac{1}{4}$ as required.
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