<s>[INST] Find $\lim_{n \to \infty} n^2 \int_{n}^{5n}\frac{x^3}{1+x^6}dx$ Question:Find the limit $\lim_{n \to \infty} n^2 \int_{n}^{5n}\frac{x^3}{1+x^6}dx$
I tried to convert it into $\frac{0}{0}$ indeterminate form,then applying 
L'Hospital's rule but the expressions in numerator are not nice to integrate.I do not know other way to solve this limit.
Can anybody help me out!
 [/INST] Since you already received good answers for the limit itself, let me show how we could have the partial terms.
$$\frac{x^3}{x^6+1}=\frac{x^3}{(x^3-i)(x^3+i)}$$ Using partial fraction decomposition
$$\frac{x^3}{x^6+1}=\frac{x-2 i}{6 \left(x^2-i x-1\right)}+\frac{x+2 i}{6 \left(x^2+i
   x-1\right)}-\frac{1}{6 (x-i)}-\frac{1}{6 (x+i)}$$ and the integration does not make much problems. Skipping the steps and recombining to have a compact result
$$12\int\frac{x^3}{x^6+1} dx=-2 \log \left(x^2+1\right)+\log \left(x^2-\sqrt{3}
   x+1\right)+\log \left(x^2+\sqrt{3} x+1\right)-$$ $$2 \sqrt{3} \tan
   ^{-1}\left(\sqrt{3}-2 x\right)-2 \sqrt{3} \tan ^{-1}\left(2
   x+\sqrt{3}\right)$$ Now, computing
$$I_k=\int_n^{kn}\frac{x^3}{x^6+1} dx \qquad \text{with} \qquad k >1$$ and expanding the result as series for large values of $n$
$$I_k=\sum_{p=0}^\infty (-1)^p\frac{1- k^{-(2+6p)}}{(2+6p)\,n^{2+6p}}$$
$$n^2 \,I_k=\sum_{p=0}^\infty (-1)^p\frac{1- k^{-(2+6p)}}{(2+6p)\,n^{6p}}$$
$$\lim_{n \to \infty} n^2 \int_{n}^{kn}\frac{x^3}{1+x^6}dx=\frac{1}{2} \left(1-\frac{1}{k^2}\right)$$ and the asymptotics
$$n^2 \,I_k=\frac{1}{2} \left(1-\frac{1}{k^2}\right)-\frac 18\left(1-\frac{1}{k^8}\right)\frac 1 {n^6}+O\left(\frac{1}{n^{12}}\right)$$
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