https://www.wolframalpha.com/input?i=∫+%281%2B1%2Fx%5E4%29%5E0.5+dx+

Quote:ThatDonGuyFirst thought: rewrite it as (sqrt(x^4 + 1) / x^2) dx, then see if you can integrate it by parts

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Thanks for your reply. I agree that we can rewrite it to (1+x^4)^0.5 / x^2

The hard part is (1+x^4)^0.5, which I can't figure out.

I would appreciate if someone could show me the solution step by step.

Quote:ssho88Quote:ThatDonGuyFirst thought: rewrite it as (sqrt(x^4 + 1) / x^2) dx, then see if you can integrate it by parts

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Thanks for your reply. I agree that we can rewrite it to (1+x^4)^0.5 / x^2

The hard part is (1+x^4)^0.5, which I can't figure out.

I would appreciate if someone could show me the solution step by step.

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I don't have time to work this out,, but let me suggest the following: When using integration by parts you need to find the integral of (1+x^4)^0.5. Try setting x^2 = tan y with (2x dx ) =sec^2(y)dy. Thus, (1+x^4)^0.5 dx becomes sqrt(1 + tan^2(y))* (1/2*tan

^{2}(y))/sec^2(y)dy.

However, (1+tan^2(y)) = sec^2 (y) so sqrt(1 + tan^2(y))*(1/2*tan

^{2}(y))/sec^2(y)dy =

1/2*tan

^{-2}(y)dy. That should be integrable, I imagine.

I hope I didn't make a mistake . . . .

You mentioned : "Thus, (1+x^4)^0.5 dx becomes sqrt(1 + tan^2(y))* (1/2*tan2(y))/sec^2(y)dy."

With substitution x^2 = tan(y), I get :-

(2x dx ) =sec^2(y)dy, and x = sqrt(tan(y))

dx = (1/2) * (1/x) * sec^2(y)dy

(1+x^4)^0.5 dx = sqrt(1 + tan^2(y)) * (1/2) * (1/x) * sec^2(y)dy

= sqrt(1 + tan^2(y)) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy

= sec(y) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy

= sec^3(y) * (1/2) * (1/sqrt(tan(y))) dy

= ????

It does not have an elementary anti-derivative.

Quote:ssho88Thanks gordonm88,

You mentioned : "Thus, (1+x^4)^0.5 dx becomes sqrt(1 + tan^2(y))* (1/2*tan2(y))/sec^2(y)dy."

With substitution x^2 = tan(y), I get :-

(2x dx ) =sec^2(y)dy, and x = sqrt(tan(y))

dx = (1/2) * (1/x) * sec^2(y)dy

(1+x^4)^0.5 dx = sqrt(1 + tan^2(y)) * (1/2) * (1/x) * sec^2(y)dy

= sqrt(1 + tan^2(y)) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy

= sec(y) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy

= sec^3(y) * (1/2) * (1/sqrt(tan(y))) dy

= ????

It does not have an elementary anti-derivative.

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yes, sorry, I typed that out quick before running out the door. I've been gone all day, until this moment. The basic idea with the substitution x^2 = tan y was to reduce sqrt(1 + x^4) to a simple trigonometric expression, while accepting whatever you get for dx. I guess it didn't work out but I don't know any other way of attacking an integral with a term sqrt(1 + x

^{2n}).

Can we pursue an antiderivative of X = sec^3(y) * (1/2) * (1/sqrt(tan(y))) by using sec^2(y) =tan^2(y) + 1 ?

X = 0.5*sec(y) *(tan^2(y) +1)/tan^(-0.5)(y) = 0.5* sec(y)*tan^(1.5)(y) +0.5*sec(y)*tan^(-0.5)(y)

This form of X might possibly have an anti-derivative with an analytic form. It is a sum of two terms that are both of the form sec x * tan

^{a}.

Only a very few functions have a closed form for their anti-derivative expressible in elementary functions. I have no idea if what you've posted is such a function, but it smells like one to me. Wolfram Alpha says the anti-derivative needs a "hypergeometric function."Quote:ssho88∫ (1+1/x^4)^0.5 dx = ?

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y=1/x, dy/dx= -1/x^2

(ds)^2 = (dx)^2 + (dy)^2

ds = dx * (1+(dy/dx)^2)^0.5 = (1 + 1/x^4)^0.5 dx

s = ∫ (1+1/x^4)^0.5 dx .

Then I got stuck here.

ETA: Wait a minute… The “length of curve, f(x) = 1/x”. Without citing boundaries (boundless), wouldn’t that be infinite?