The limit sequence given by integral I have problem with calculating a limit of
$$\lim_{n\rightarrow\infty}\left(n^2\int_n^{3n}\frac{x+1}{x^4+1} \ dx\right)$$
without calculating integral 
$\int_n^{3n}\frac{x+1}{x^4+1} \ dx$
I tried the mean value theorem for integrals (IMVT) but it didn't work 
 A: Hint:
You may squeeze the integral as follows:
$$\int_n^{3n}\frac{1}{(x+1)^3} \ dx= \int_n^{3n}\frac{x+1}{(x+1)^4} \ dx\leq \int_n^{3n}\frac{x+1}{x^4+1} \ dx \leq  \int_n^{3n}\frac{x+1}{x^4} \;dx = \int_n^{3n}\left(\frac{1}{x^3}+ \frac{1}{x^4}\right) \;dx $$
Multiplying the results on both sides by $n^2$ and taking the limit gives $\frac{4}{9}$.
A: You can apply L'Hôpital's rule to the rearranged indeterminate form of the limit as follows
$$\begin{align}
\lim_{n\to\infty}\frac{\int_{n}^{3n}\frac{x+1}{x^4+1}\mathrm{d}x}{1/n^2}
&=\lim_{n\to\infty}\frac{3\left(\frac{3n+1}{81n^4+1}\right)-\frac{n+1}{n^4+1}}{-2/n^3}\\
&=-\frac12\lim_{n\to\infty}\left(\frac{3n^3(3n+1)}{81n^4+1}-\frac{n^3(n+1)}{n^4+1}\right)\\
&=-\frac12\left(\frac19-1\right)\\
&=\frac49\\
\end{align}$$
A: Putting $x=nt$ in the integral we get the expression under limit as $$n^3\int_{1}^{3}\frac{nt+1}{n^4t^4+1}\,dt$$ and we can take limit inside integral to get $\int_{1}^{3}t^{-3}\,dt=4/9$.
The justification of taking limit under integral sign is easily given by considering the difference $$f(t) =\frac{n^4t+n^3}{n^4t^4+1}-\frac{1}{t^3}$$ which is bounded by $1/n$.
