Surface Area of Rotated Curve: How to Solve for the Integral of sqrt(x^4+1)/x^3?

[freshman2013]

Homework Statement

This is actually part of this problem: find the surface area of (1/x) rotated about the x-axis from 1 to infinity

Homework Equations

Surface Area=∫2\piy\sqrt{1+(dy/dx)^2}

The Attempt at a Solution

I took the derivative of (1/x)=-1/x^2
so I have ∫2\pi(1/x)\sqrt{1+1/x^4}
=∫2\pi(1/x)\sqrt{(x^4+1)/x^4}
and got 2\pi\sqrt{(x^4+1)}/x^3
Now I'm stuck. I can't do trig sub since they aren't powers of 2, x^4+1 isn't a perfect square, so what should I do now?

 

[Office_Shredder]

Instead of trying to evaluate it directly you might want to first consider heuristically what simpler function that integral looks like when x is very large (I realize this sounds like a weird suggestion but it will make sense in the end)

 

[freshman2013]

Ok, it "looks" like 1/x as x approaches infinity. How will this help me?

 

[Office_Shredder]

Well, can you tell me what
\int_{1}^{\infty} \frac{1}{x} dx
is?

 

[freshman2013]

infinity, right?

EDIT, so the original has also to be infinity?

 

[Office_Shredder]

Right, now can you use that result to show that your integral is infinity also? Not in a 'it looks like the same kind of' way but in stating correct mathematical theorems.

 

[freshman2013]

No, I never seen nor done your method before
EDIT: wait, limit comparison test? Even though I only learned of it for series, it's the only thing I can think of that makes sense

 

[Office_Shredder]

Oh, that's unfortunate because I have never seen an example of this question where someone expects you to actually evaluate the integral.

Yes, there is a version of the limit comparison test for integrals I had assumed you were supposed to do something like that for this integral (although just the regular comparison test is taught more often, anything that involves comparing integrals is sufficient).

As far as calculating it exactly, I highly recommend the substitution u=1/x⁴ but be prepared for a solid slog through the muck on this one.