ssho88 Joined: Oct 16, 2011
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July 27th, 2022 at 2:40:49 AM permalink
∫ (1+1/x^4)^0.5 dx = ?
unJon Joined: Jul 1, 2018
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Thanks for this post from: 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 Joined: Jun 22, 2011
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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 Joined: Oct 16, 2011
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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

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 Joined: Feb 18, 2015
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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

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.

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 Joined: Oct 16, 2011
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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 Joined: Feb 18, 2015
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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.

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 Joined: Oct 19, 2009
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July 29th, 2022 at 10:04:32 AM permalink
Quote: ssho88

∫ (1+1/x^4)^0.5 dx = ?

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 Joined: Oct 16, 2011
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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 Joined: Jun 22, 2010
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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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unJon Joined: Jul 1, 2018
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July 29th, 2022 at 12:53:34 PM permalink
If you have wolfram alpha pro you can just type in arc length 1/X and click show step by step solution.
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gordonm888 Joined: Feb 18, 2015
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July 29th, 2022 at 2:38:04 PM permalink
Recognizing that this was either an unsolvable integral (i.e., no analytic form for the anti-derivative) or at least one for which the solution is not widely known, I was just typing in notes in my posts as I studied it. I realized that my posts were not directly helpful, sho88, I was just trying to come up with a good idea on how to start the analysis.

I did notice that the integral of (1/sqrt(a2+u2) = arsinh (u/a) + C This is the closest form that I have found for a simple integral relevant to your problem.

Also, the integral of(1/(u*sqrt(a2+u2)) = arcsch |(u/a)| + C where I think arcsch is an arc-hyperbolic cosecant.

All of which seems to be in sync with teliot's comment about Wolfram's advice to look at hyperbolic functions. However, hyperbolas have different mathematical properties than the curve you are studying and I'm not sure if they will reduce to your curve by, say, setting the distance between the two foci to 0.

Certainly not an "easy math problem."
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ssho88 Joined: Oct 16, 2011
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July 29th, 2022 at 6:19:44 PM permalink
Quote: gordonm888

Recognizing that this was either an unsolvable integral (i.e., no analytic form for the anti-derivative) or at least one for which the solution is not widely known, I was just typing in notes in my posts as I studied it. I realized that my posts were not directly helpful, sho88, I was just trying to come up with a good idea on how to start the analysis.

I did notice that the integral of (1/sqrt(a2+u2) = arsinh (u/a) + C This is the closest form that I have found for a simple integral relevant to your problem.

Also, the integral of(1/(u*sqrt(a2+u2)) = arcsch |(u/a)| + C where I think arcsch is an arc-hyperbolic cosecant.

All of which seems to be in sync with teliot's comment about Wolfram's advice to look at hyperbolic functions. However, hyperbolas have different mathematical properties than the curve you are studying and I'm not sure if they will reduce to your curve by, say, setting the distance between the two foci to 0.

Certainly not an "easy math problem."

Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Last edited by: ssho88 on Jul 29, 2022
gordonm888 Joined: Feb 18, 2015
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July 30th, 2022 at 1:20:57 PM permalink
Nice. But with y = x^4, is there any convergence issue when x<1?
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teliot Joined: Oct 19, 2009
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July 30th, 2022 at 5:38:26 PM permalink
Quote: ssho88

Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

For a numerical approximation, this is exactly what I would do.
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ssho88 Joined: Oct 16, 2011
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July 30th, 2022 at 7:00:38 PM permalink
Quote: gordonm888

Nice. But with y = x^4, is there any convergence issue when x<1?

Convergence issue is my concern too. Thanks for your comment.
ssho88 Joined: Oct 16, 2011
• Posts: 655
July 30th, 2022 at 7:01:58 PM permalink
Quote: teliot

Quote: ssho88

Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

For a numerical approximation, this is exactly what I would do.

gordonm888 Joined: Feb 18, 2015
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August 6th, 2022 at 5:14:38 AM permalink
Did the Taylor series converge? Were you able to solve this integral?
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ssho88 Joined: Oct 16, 2011
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August 6th, 2022 at 6:53:50 AM permalink
Here is the integral solution:- However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?
unJon Joined: Jul 1, 2018
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August 6th, 2022 at 7:09:14 AM permalink
Quote: ssho88

Here is the integral solution:- However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?

Just the same comment I made twice before.

https://www.wolframalpha.com/input?i=arc+length+1%2FX&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22a%22%7D+-%3E%221%22&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22b%22%7D+-%3E%223%22
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ssho88 Joined: Oct 16, 2011
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August 6th, 2022 at 7:33:55 AM permalink
Quote: unJon

Quote: ssho88

Here is the integral solution:- However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?

Just the same comment I made twice before.

https://www.wolframalpha.com/input?i=arc+length+1%2FX&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22a%22%7D+-%3E%221%22&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22b%22%7D+-%3E%223%22

So Taylor series is not a viable way to find the arc length of 1/x ?
unJon Joined: Jul 1, 2018
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August 6th, 2022 at 8:09:39 AM permalink
Quote: ssho88

Quote: unJon

Quote: ssho88

Here is the integral solution:- However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?

Just the same comment I made twice before.

https://www.wolframalpha.com/input?i=arc+length+1%2FX&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22a%22%7D+-%3E%221%22&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22b%22%7D+-%3E%223%22