Evaluate $\lim_{n\to\infty}nI_n$ with $I_n=\int_0^1\frac{x^n}{x^2+3x+2}dx$ We have to evaluate: $$\lim_{n\to\infty}nI_n$$ with $$I_n=\int_0^1\frac{x^n}{x^2+3x+2}\:dx.$$

There is an elegant way to solve this problem?


Here is all my steps:


*

*My first ideea was to find a recurrence relation such that:


$$I_{n+2}+3I_{n+1}+2I_n=\frac{1}{n+1},\forall x\in\mathbb{N}$$


*

*Next step I show that $\forall x\in[0,1]\Rightarrow I_{n}\ge I_{n+1}\ge I_{n+2}$ 


Therefore it involving that: $$6I_n\ge 4I_{n+1}+2I_n\ge\frac{1}{n+1},\forall x\in\mathbb{N}$$
As I said above $$6I_{n+2}\leq 4I_{n+2}+2I_n\leq\frac{1}{n+1}$$ 
$\Rightarrow \frac{n}{6(n+1)}\leq nI_n\leq\frac{n}{6(n-1)},\forall x\in\mathbb{N}$
Therefore by squeeze thereom:
$$nI_n\to\frac{1}{6}\:as\:n\to\infty$$
 A: We may just integrate by parts,
$$
\begin{align}
I_n=\int_0^1\frac{x^n}{(x+1)(x+2)}dx&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{(x+1)(x+2)}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{(2x+3)\:x^{n+1}}{(x+1)^2(x+2)^2}\:dx\\\\
&=\frac1{6(n+1)}+\frac{1}{n+1}\int_0^1\frac{(2x+3)}{(x+1)^2(x+2)^2}\:x^{n+1}dx\\\\
&=\frac1{6(n+1)}+\frac{1}{n+1}J_n \tag1
\end{align}
$$ and one may observe that
$$
0\leq \int_0^1\frac{(2x+3)}{(x+1)^2(x+2)^2}\:x^{n+1}dx\leq \frac{(2\times1+3)}{(0+1)^2(0+2)^2}\int_0^1x^{n+1}dx
$$ or
$$
0\leq J_n\leq \frac{5}{4}\frac{1}{(n+2)}. \tag2
$$
Then using $(1)$ and $(2)$ gives easily 

$$ \lim_{n \to +\infty}nI_n=\frac16.$$

A: Substitute $x\mapsto x^{1/(n+1)}$ and use Dominated Convergence:
$$
\begin{align}
n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x
&=\frac{n}{n+1}\int_0^1\frac1{x^2+3x+2}\,\mathrm{d}x^{n+1}\\
&=\frac{n}{n+1}\int_0^1\frac1{x^{2/(n+1)}+3x^{1/(n+1)}+2}\,\mathrm{d}x\\
&\to1\int_0^1\frac1{1+3\cdot1+2}\,\mathrm{d}x\\
&=\frac16
\end{align}
$$

A More Basic Approach
$$
\begin{align}
\frac16-\left(n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x\right)
&=\int_0^1\left(\frac16-\frac{x^{1/n}}{x^{2/n}+3x^{1/n}+2}\right)\,\mathrm{d}x\tag1\\
&=\frac16\int_0^1\left(\frac{x^{2/n}-3x^{1/n}+2}{x^{2/n}+3x^{1/n}+2}\right)\,\mathrm{d}x\tag2\\
&\le\frac1{12}\int_0^1\left(x^{2/n}-3x^{1/n}+2\right)\,\mathrm{d}x\tag3\\[3pt]
&=\frac1{12}\left(\frac{n}{n+2}-3\frac{n}{n+1}+2\right)\tag4\\[6pt]
&=\frac{7n-2}{12(n+1)(n+2)}\tag5
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x^{1/n}$ then bring the $\frac16$ inside the integral
$(2)$: algebra; $x^{2/n}-3x^{1/n}+2=\left(x^{1/n}-1\right)\left(x^{1/n}-2\right)\ge0$ on $[0,1]$
$(3)$: since the integrand is positive, bound it by replacing the denominator with its minimum
$(4)$: integrate
$(5)$: simpllfy
Therefore,
$$
\frac16-\frac{7n-2}{12(n+1)(n+2)}\le\left(n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x\right)\le\frac16\tag6
$$
Apply the Squeeze Theorem to get
$$
\lim_{n\to\infty}n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x=\frac16\tag7
$$
