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

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

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

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

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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 $$ \lim_{n\to\infty}n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x=\frac16\tag7 $$