Evaluate: $\lim_{n\to\infty} n^3\int_n^{2n} \frac x{1+x^5}\, dx. $ Is there a way to evaluate: $$\lim_{n\to\infty}n^3\int_n^{2n}\frac x{1+x^5}\, dx,$$ without evaluating the indefinite integral: $$\int \frac x{1+x^5}\, dx\;?$$ Please suggest. Thanks in advance. 
 A: Put $y=\frac  x n$. You get $n^{3} \int_1^{2} \frac {ny} {1+n^{5}y^{5}} ndy$ or $ \int_1^{2} \frac {n^{5}y} {1+n^{5}y^{5}}dy$. The integrand tends to $y^{-4}$ and it is bounded by $1$. By DCT the limit is $\int_1^{2}y^{-4} dy=\frac  7 {24}$.
A: You can use l'Hospital as follows:
$$t^3\int_t^{2t}\frac x{1+x^5}dx = \frac{\int_t^{2t}\frac x{1+x^5}dx}{t^{-3}}$$ $$\stackrel{L'Hosp.}{\sim}\frac{2\frac{2t}{1+(2t)^5} - \frac{t}{1+t^5}}{-3t^{-4}}=-\frac 13 \left(\frac{4}{\frac 1{t^5}+32}-\frac 1{\frac 1{t^5}+1}\right)\stackrel{t\to+\infty}{\longrightarrow}\frac 7{24}$$
Note that for $t$ large enough the integrand is decreasing and you have $\int_t^{2t}\frac x{1+x^5}dx \leq t \frac t{1+t^5}\leq \frac 1{t^3}\stackrel{t\to\infty}{\longrightarrow}0$. This justifies the use of L'Hospital here.
A: This is mainly an elaboration on Kavi Rama Murthy's answer, in case you don't have DCT at your disposal.
Letting $x=nu$, we have
$$\begin{align}
n^3\int_n^{2n}{x\over1+x^5}dx
&=\int_1^2{n^5u\over1+n^5u^5}du\\
&=\int_1^2{u\over u^5+(1/n^5)}du\\
&=\int_1^2{du\over u^4}+\int_1^2\left({u\over u^5+(1/n^5)}-{1\over u^4} \right)du\\
&={7\over24}-{1\over n^5}\int_1^2{du\over u^4(u^5+(1/n^5))}
\end{align}$$
and
$$0\lt{1\over n^5}\int_1^2{du\over u^4(u^5+(1/n^5))}\lt{1\over n^5}\int_1^2{du\over u^9}\to0$$
A: Let's put $x=u^{-1/3}$ in the integral to get the expression under limit as $$\frac{n^3}{3}\int_{1/8n^3}^{1/n^3}\frac{du}{1+u^{5/3}}$$ and we can  rewrite it further as $$\frac{1}{3}\left(\frac{1}{1/n^3}\int_{0}^{1/n^3}\frac{du} {1+u^{5/3}}-\frac{1}{8}\cdot\frac{1}{1/8n^3}\int_{0}^{1/8n^3}\frac{du}{1+u^{5/3}}\right)$$ Now using fundamental theorem of calculus the desired limit is $$\frac{1}{3}\left(1-\frac{1}{8}\right)=\frac{7}{24}$$
A: Under $x^{-3}\to x$,
$$ \frac{1}{n^3}\int_{n}^{2n}\frac{x}{1+x^5}dx=\frac{n^3}{3}\int_{\frac1{8n^3}}^{\frac1{n^3}}\frac{dx}{1+x^{\frac{5}{3}}}. $$
By the MVT for integrals, there is $c_n\in (\frac1{8n^3},\frac1{n^3})$ such that
$$ \frac{1}{n^3}\int_{n}^{2n}\frac{x}{1+x^5}dx=\frac1{n^3}\bigg(\frac1{8n^3}-\frac1{n^3}\bigg)\frac{1}{1+c_n^{\frac{5}{3}}}=\frac{7}{24}. $$
So
$$ \lim_{n\to\infty}\frac{1}{n^3}\int_{n}^{2n}\frac{x}{1+x^5}dx=\lim_{n\to\infty}\frac{7}{24}\frac{1}{1+c_n^{\frac{5}{3}}}=\frac{7}{24}. $$
