Integral involving inverse of a function If $f(x)=x^3-x^2+x$, evaluate $$\lim_{n \to \infty} \int _{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)}$$ I substituted $x=f(t)$  but was unable to convert it back to $x$ and I think there would be a better approach. Some hints please. Thanks.
 A: I won't go into details, but try to follow the logic on the argumentation.
Look at this limit for a polynomial that grows at $\mathcal{O}(n^3)$. If we are speaking about its inverse, you shall agree that $x^3$ is much more relevant than the other terms, no? So, if we have we can write, for large $y$, $$f^{-1}(y) = \sqrt[3]{y} + \mathcal{o}(y^{2/3})$$
So, your integration, can be largely seen as
$$\lim_{n \to \infty} \int _{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)} = \lim_{n \to \infty} \int _{n}^{2n}\frac{dx}{x+\sqrt[3]{x}} = \lim_{n \to \infty} \log \left(\left[\frac{2^{2/3} n^{2/3}+1}{n^{2/3}+1}\right]^{3/2} \right)=\log(2)$$
A: For any $x>1$ we have $\left(x-\frac{1}{3}\right)^3 \leq f(x) \leq x^3 $, hence $f^{-1}(2n)-f^{-1}(n)$ is $O(\sqrt[3]{n})$ for large $n$ and $\frac{f^{-1}(2n)}{f^{-1}(n)}\to \sqrt[3]{2}$ as $n\to +\infty$
Moreover,
$$ \int \frac{3x^2-2x+1}{x^3+x}\,dx = C-2\arctan(x)+\log(x)+\log(1+x^2)\tag{1}$$
hence we may deal with
$$ \int_{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)} = \int_{f^{-1}(n)}^{f^{-1}(2n)}\frac{3x^2-2x+1}{x^3+x}\,dx \tag{2}$$
through the identities:
$$ \arctan(Cx)-\arctan(x) = \left(1-\frac{1}{C}\right)\frac{1}{x}+O\left(\frac{1}{x^2}\right), $$
$$ \log(Cx)-\log(x) = \log C, $$
$$ \log(1+C^2 x^2)-\log(1+x^2) = 2\log C+O\left(\frac{1}{x^2}\right)\tag{3}$$
with $C=\sqrt[3]{2}$. It follows that:

$$ \lim_{n\to +\infty}\int_{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)} = 3\log C = \color{red}{\log 2}.\tag{4}$$

A simpler approach comes from noticing that $f^{-1}(x)^3+f^{-1}(x)=x+f^{-1}(x)^2$. 
Since $f^{-1}(x)^2 = o(n)$ for any $x\in[n,2n]$,
$$ \lim_{n\to +\infty}\int_{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)} = \lim_{n\to +\infty}\int_{n}^{2n}\frac{dx}{x+f^{-1}(x)^2} = \lim_{n\to +\infty}\int_{n}^{2n}\frac{dx}{x}=\color{red}{\log 2}.$$
A: Note that $f(u)\approx u^3$ for large $u$. So $\lim_{u\to\infty}\frac{f(u)}{u^3}=1$ or $\lim_{x\to\infty}\frac{x}{(f^{-1}(x))^3}=1$ and $\lim_{x\to\infty}\frac{f^{-1}(x))}{x}=0$. Thus for $x>1$
$$ \lim_{n\to\infty}\frac{(f^{-1}(nx))^3}{n}=x,\lim_{n\to\infty}\frac{f^{-1}(nx)}{n}=0. $$
Therefore
\begin{eqnarray}
&&\lim_{n \to \infty} \int _{n}^{2n}\frac{dx}{(f^{-1}(x))^3+f^{-1}(x)} \\
&=& \lim_{n \to \infty} \int _{1}^{2}\frac{ndx}{(f^{-1}(nx))^3+f^{-1}(nx)}\\
&=& \lim_{n \to \infty} \int _{1}^{2}\frac{dx}{\frac{(f^{-1}(nx))^3}{n}+\frac{f^{-1}(nx)}{n}}\\
&=& \int _{1}^{2}\frac{1}{x}dx\\\
&=&\log(2)
\end{eqnarray}
