The intergral of Sqrt[x^3+1] is enough to drive a man insane

[bishnu]

I have been trying to find the intergral of [math]Int{Sqrt[x^3+1]}[/math]... maybe you guys can help me. First of all im defining an unknown function M(x) this function has the property of M(x)^3+1=1/M'(x)^2 so by subsituting this function in to the intergral i get the intergral to be equal to the inverse of M(x). Now this really doesnt help much because to solve the differntal equation is in essence just solving the intergral unless i some how am able genterate a series solution which i dont think would be possible and it wouldnt be useful either because i would have to take the inverse of that function anyway. So my only option now is to explore the properties of the function M(x). OKay M^-1(x)' is equal to Sqrt(X^3+1)...well that really doesnt help. Damn it what else can i do...i think the best choice would be a series solution now and then invert it...does anyone have any ideas

[VikingF]

http://integrals.wolfram.com/

 

Try this.

[Dave]

Yes, the view of "EllipticF" signifies that this isn't exactly an easy integral

[Johnny5]

I have been trying to find the intergral of [math]Int{Sqrt[x^3+1]}[/math']... maybe you guys can help me. First of all im defining an unknown function M(x) this function has the property of M(x)^3+1=1/M'(x)^2 so by subsituting this function in to the intergral i get the intergral to be equal to the inverse of M(x). Now this really doesnt help much because to solve the differntal equation is in essence just solving the intergral unless i some how am able genterate a series solution which i dont think would be possible and it wouldnt be useful either because i would have to take the inverse of that function anyway. So my only option now is to explore the properties of the function M(x). OKay M^-1(x)' is equal to Sqrt(X^3+1)...well that really doesnt help. Damn it what else can i do...i think the best choice would be a series solution now and then invert it...does anyone have any ideas

 

The integral is absolutely trivial. What you need is the series expansion of:

 

 

[math] (1+U)^\alpha [/math]

 

Where U is a function of x, that is U=U(x).

 

I can help you later.

 

In your problem, alpha = 1/2, and U(x) = x^3.

 

Your final answer will be a series, and if you can put it into a program to rapidly do the addition for you.

 

If you want an answer written in terms of elementary functions, then you will either need to use a different approach, or figure out a way to transform the final series which you got after integrating, back into elementary functions. This isn't always possible though.

 

Regards

[Dave]

What?

 

I can assure you that the answer is most certainly not trivial. Just a cursory glance at what Mathematica gives will reveal this much. Another 10 minutes working on the problem will confirm it

[Johnny5]

What?

 

I can assure you that the answer is most certainly not trivial. Just a cursory glance at what Mathematica gives will reveal this much. Another 10 minutes working on the problem will confirm it

 

The addition takes awhile, but I made an improvement on the binomial expansion formula.

[Dave]

Yeah, the addition would take quite a long time. An infinitely long time in fact - it's a non-integer power.

[Meir Achuz]

That answer is typical of Mathematica. It "solves" your relatively simple integral in terms of a complicated function of a more complicated (elliptic) integral.

Your integral can be readily evaluated by expanding the sqrt in a binomial series.

The non-integral power gives no difficulty in summing a binomial series.

[Dave]

Fair enough. I just don't have time to go through every single integration (I have my own exams to worry about). I've also never seen that particular method of integration so I stand corrected. However, I do stand by the fact that the integral is not trivial