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!

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
$$n^2 \,I_k=\frac{1}{2} \left(1-\frac{1}{k^2}\right)-\frac 18\left(1-\frac{1}{k^8}\right)\frac 1 {n^6}+O\left(\frac{1}{n^{12}}\right)$$
