ssho88
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July 27th, 2022 at 2:40:49 AM permalink
∫ (1+1/x^4)^0.5 dx = ?
unJon
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Thanks for this post from:
BleedingChipsSlowly
July 27th, 2022 at 3:32:34 AM permalink
I would copy it into Wolfram Alpha.

https://www.wolframalpha.com/input?i=∫+%281%2B1%2Fx%5E4%29%5E0.5+dx+
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ThatDonGuy
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July 27th, 2022 at 4:31:07 PM permalink
First thought: rewrite it as (sqrt(x^4 + 1) / x^2) dx, then see if you can integrate it by parts
ssho88
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July 27th, 2022 at 6:12:55 PM permalink
Quote: ThatDonGuy

First thought: rewrite it as (sqrt(x^4 + 1) / x^2) dx, then see if you can integrate it by parts
link to original post



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.
gordonm888
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July 28th, 2022 at 2:40:54 PM permalink
Quote: ssho88

Quote: ThatDonGuy

First thought: rewrite it as (sqrt(x^4 + 1) / x^2) dx, then see if you can integrate it by parts
link to original post



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.
link to original post



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*tan2(y))/sec^2(y)dy.

However, (1+tan^2(y)) = sec^2 (y) so sqrt(1 + tan^2(y))*(1/2*tan2(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 . . . .
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ssho88
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July 28th, 2022 at 7:44:48 PM permalink
Thanks 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.
Last edited by: ssho88 on Jul 28, 2022
gordonm888
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July 28th, 2022 at 9:59:31 PM permalink
Quote: ssho88

Thanks 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.
link to original post



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 + x2n).

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 * tana.
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teliot
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July 29th, 2022 at 10:04:32 AM permalink
Quote: ssho88

∫ (1+1/x^4)^0.5 dx = ?
link to original post

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."
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ssho88
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July 29th, 2022 at 10:52:24 AM permalink
My actual intention is to find the length of curve, f(x) = 1/x, the expression ∫ (1+1/x^4)^0.5 dx was derived as below :-

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.
camapl
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July 29th, 2022 at 11:49:40 AM permalink
I don’t know if you’re expected to provide an accurate answer, but aren’t there tables with approximate answers for certain “difficult” functions?

ETA: Wait a minute… The “length of curve, f(x) = 1/x”. Without citing boundaries (boundless), wouldn’t that be infinite?
Last edited by: camapl on Jul 29, 2022
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