text: # 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?
  
  -
  Do you know about the Dominated Convergence Theorem? Your problem is a simple application of it. – nayrb Jul 10 '13 at 18:04
  
  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 $$\lim_n\int_0^1\frac{nx}{nx^3+1}=\infty.$$
  
  -
  No $dx$? It's all wrong! – Git Gud Jul 10 '13 at 18:10
  I'll put the $dx$ if it makes you happy... ;) – Martin Argerami Jul 10 '13 at 18:10
  $\ddot \smile{{}}$ – Git Gud Jul 10 '13 at 18:11
  I think i got it, Thanks. – StationaryTraveller Jul 10 '13 at 18:13
  
  Not as good, as other answers, but still...
  
  Since $\ 0<x<1$$\int\limits_0^1\frac{nx}{nx^3+1}dx>\int\limits_0^1\frac{nx}{nx^2+1}dx=\ln\sqrt{n+1}\to\infty$$ - Actually, this one is great. – StationaryTraveller Jul 10 '13 at 19:31 I like this one better than mine. – Martin Argerami Jul 11 '13 at 4:26 Thanks. Others are just more obvious and give general idea of how to handle these types of problems, so they are more useful. Mine is not so obvious. But I can see why you like mine more - it's somewhat shorter and simplier ) Sorry for possible language mistakes, I'm Russian. – George Stobbart Jul 11 '13 at 9:22 Step 1: $$\frac{nx}{nx^3+1}=\frac{x}{x^3+\frac{1}{n}}$$ Step 2: $$\int_0^1 \frac{x}{x^3}dx~~\textrm{ diverges}$$ - so here do we need to show uniform Convergence and then take limits inside? – Neeraj Bhauryal Jul 10 '13 at 18:08 So you first calculated the limit? Is it mathematically correct? – StationaryTraveller Jul 10 '13 at 18:08 This isn't a complete solution, just a pair of hints. See Martin's answer for a complete treatment. – vadim123 Jul 10 '13 at 18:09 I got it, Thank you. – StationaryTraveller Jul 10 '13 at 18:13 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. - @MartinArgerami sorry, my bad. Will correct in a sed. – gt6989b Jul 10 '13 at 18:14 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.
  
  -
explanation: The document contains a math question and multiple answers to this question. The question is about evaluating the limit of an integral, and the answers provide different approaches to solving it, including the use of the Dominated Convergence Theorem and other methods.
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distilabel_metadata.content: # 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?
  
  -
  Do you know about the Dominated Convergence Theorem? Your problem is a simple application of it. – nayrb Jul 10 '13 at 18:04
  
  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 $$\lim_n\int_0^1\frac{nx}{nx^3+1}=\infty.$$
  
  -
  No $dx$? It's all wrong! – Git Gud Jul 10 '13 at 18:10
  I'll put the $dx$ if it makes you happy... ;) – Martin Argerami Jul 10 '13 at 18:10
  $\ddot \smile{{}}$ – Git Gud Jul 10 '13 at 18:11
  I think i got it, Thanks. – StationaryTraveller Jul 10 '13 at 18:13
  
  Not as good, as other answers, but still...
  
  Since $\ 0<x<1$$\int\limits_0^1\frac{nx}{nx^3+1}dx>\int\limits_0^1\frac{nx}{nx^2+1}dx=\ln\sqrt{n+1}\to\infty$$ - Actually, this one is great. – StationaryTraveller Jul 10 '13 at 19:31 I like this one better than mine. – Martin Argerami Jul 11 '13 at 4:26 Thanks. Others are just more obvious and give general idea of how to handle these types of problems, so they are more useful. Mine is not so obvious. But I can see why you like mine more - it's somewhat shorter and simplier ) Sorry for possible language mistakes, I'm Russian. – George Stobbart Jul 11 '13 at 9:22 Step 1: $$\frac{nx}{nx^3+1}=\frac{x}{x^3+\frac{1}{n}}$$ Step 2: $$\int_0^1 \frac{x}{x^3}dx~~\textrm{ diverges}$$ - so here do we need to show uniform Convergence and then take limits inside? – Neeraj Bhauryal Jul 10 '13 at 18:08 So you first calculated the limit? Is it mathematically correct? – StationaryTraveller Jul 10 '13 at 18:08 This isn't a complete solution, just a pair of hints. See Martin's answer for a complete treatment. – vadim123 Jul 10 '13 at 18:09 I got it, Thank you. – StationaryTraveller Jul 10 '13 at 18:13 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. - @MartinArgerami sorry, my bad. Will correct in a sed. – gt6989b Jul 10 '13 at 18:14 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.
  
  -
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distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and multiple answers to this question. The question is about evaluating the limit of an integral, and the answers provide different approaches to solving it, including the use of the Dominated Convergence Theorem and other methods.
  
  Label: question-and-answer