Evaluate the integral: $\lim \limits_{n\to\infty}\int_0^1\frac{nx}{nx^3+1}$ Evaluate the integral:
$$\lim \limits_{n\to\infty}\int_0^1\frac{nx}{nx^3+1}dx$$
I'm pretty much stuck on how to solve this one:
$$\int_0^1\frac{nx}{nx^3+1}dx$$
or even getting the improper integral.
What can i do?
 A: Not as good, as other answers, but still...
Since $\ 0<x<1$$$\int\limits_0^1\frac{nx}{nx^3+1}dx>\int\limits_0^1\frac{nx}{nx^2+1}dx=\ln\sqrt{n+1}\to\infty$$
A: If you think that your integral is $\int_0^1\frac{x}{x^3+1/n}$, in the limit it is the integral of $1/x^2$, so you should expect it to diverge. 
Then you can do the following:
$$
\int_0^1\frac{nx}{nx^3+1}=\int_0^1\frac{x}{x^3+1/n}\geq\int_{1/n^{1/3}}^1\frac{x}{x^3+1/n}\geq\int_{1/n^{1/3}}^1\frac{x}{2x^3}\\ \ \\=\int_{1/n^{1/3}}^1\frac1{x^2}=n^{1/3}-1.
$$
So $$\lim_n\int_0^1\frac{nx}{nx^3+1}=\infty.$$
A: Step 1: $$\frac{nx}{nx^3+1}=\frac{x}{x^3+\frac{1}{n}}$$
Step 2: $$\int_0^1 \frac{x}{x^3}dx~~\textrm{ diverges}$$
A: To add it in more explicit language:
You would like to exchange the limit and the integral, i.e. to write
$$\lim_{n \to \infty} \int_0^1 f_n(x) dx
  = \int_0^1 \lim_{n \to \infty} f_n(x) dx
  = \int_0^1 \frac{dx}{x^2} = +\infty.$$
The Lebesgue Monotone Convergence Theorem allows to do just that.
A: I'd try to split this integral using partial fractions, treating $n$ as a constant :
$$\begin{eqnarray} I &=& \int_0^1 \frac{nx}{n x^3 + 1} dx \\
&=& \int_0^1 \frac A{n^{1/3} x + 1} dx + \int_0^1 \frac {B x + C}{n^{2/3} x^2 - n^{1/3}x + 1} dx \end{eqnarray}$$
Both integrals can be done using classical techniques. The rest is up to you.
