# 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$\pi$y$\sqrt{1+(dy/dx)^2}$

## 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????

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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?

Staff Emeritus
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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?

Staff Emeritus
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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

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