Integral: sqrt(1+((x^4-1)/(2x^2))^2)

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Discussion Overview

The discussion revolves around solving the integral of the expression sqrt(1+((x^4-1)/(2x^2))^2). Participants explore various methods of simplifying the expression and discuss steps toward integration without reaching a consensus on the final approach.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant requests help with the integral, expressing uncertainty about how to solve it.
  • Another participant suggests expanding the expression inside the square root and simplifying it, prompting further calculations.
  • There is a correction regarding the expansion of (x^4-1)^2, with a participant noting the correct form as x^8-2x^4+1.
  • Participants discuss the factorization of the resulting expression, with one noting that it can be factored easily.
  • There is a suggestion to substitute x^4 with y to simplify the problem further.
  • One participant expresses confusion about the next steps after simplifying the expression to √(x^4+1)²/2x².
  • Another participant breaks down the integral into simpler components, indicating a progression toward a solution.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final steps for solving the integral, and there are multiple approaches and corrections discussed throughout the thread.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the simplification and integration process, which may affect the clarity of the discussion.

Alexx1
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http://integrals.wolfram.com/index.jsp?expr=sqrt(1%2B((x^4-1)%2F(2x^2))^2)&random=false

Can someone explain me how to solve an integral like this?
I tried it, but I have absolutely no idea how to solve it..
 
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Welcome to PF!

Hi Alexx1! Welcome to PF! :smile:
sqrt(1+((x^4-1)/(2x^2))^2)
Alexx1 said:
http://integrals.wolfram.com/index.jsp?expr=sqrt(1%2B((x^4-1)%2F(2x^2))^2)&random=false

Can someone explain me how to solve an integral like this?
I tried it, but I have absolutely no idea how to solve it..

Do you mean \sqrt(1+(\frac{x^4-1}{2x^2})^2) ? :confused:
 


tiny-tim said:
Hi Alexx1! Welcome to PF! :smile:



Do you mean \sqrt(1+(\frac{x^4-1}{2x^2})^2) ? :confused:

Yes, it's that integral ;)
 
Alexx1 said:
Yes, it's that integral ;)

ok, then (without integrating) expand everything inside the √, and then simplify …

what do you get? :smile:​
 
tiny-tim said:
ok, then (without integrating) expand everything inside the √, and then simplify …

what do you get? :smile:​

(8x^8+2x^4+x)/(4x^4)
 
(try using the X2 tag just above the Reply box :wink:)
Alexx1 said:
(8x^8+2x^4+x)/(4x^4)

uhhh? :confused: not even close :redface:
 
tiny-tim said:
(try using the X2 tag just above the Reply box :wink:)


uhhh? :confused: not even close :redface:


((x^4-1)/2x^2)^2) = ((x^4-1)^2)/((2x^2)^2)

= (x^8+1-2x^4)/(4x^4)

==> 1 + (x^8+1-2x^4)/(4x^4) = (4x^4+x^8+1-2x^4)/(4x^4)

= (x^8+2x^4+1)/(4x^4)

This is the correct answer?
I made a mistake it had to be: x^8 (instead of 8x^8) and 1 (instead of x)
 
(please use the X2 tag just above the Reply box … it makes it much easier to read)
Alexx1 said:
… (x^8+2x^4+1)/(4x^4)

ok! :smile:

and the square-root of that is … ?​
 
What should've made factorising easy to spot is that you first expanded
(x^4-1)^2=x^8-2x^4+1

and after adding 4x^4 you now have
x^8+2x^4+1

Isn't it clear how this can be factored?
 
  • #10
The crucial point is that 4a+ (x- a)^2= 4a+ x^2- 2ax+ a^2= x^2+ 2ax+ a^2= (x+a)^2.

That's used a lot to produce easy "arc length" problems!
 
  • #11
tiny-tim said:
(please use the X2 tag just above the Reply box … it makes it much easier to read)ok! :smile:

and the square-root of that is … ?​


= √x8+2x4+1/√4x4

= √(x4+1)2/2x2

= x4+1/2x2
 
  • #12
Alexx1 said:
But now I don't know what to do with : √x8+2x4+1

come on, think! :smile:

(or put x4 = y :wink:)​
 
  • #13
tiny-tim said:
come on, think! :smile:

(or put x4 = y :wink:)​


Ok, I found it!

= √x8+2x4+1/√4x4

= √(x4+1)2/2x2

= x4+1/2x2

==> Integral x4+1/2x2

= (1/2) Integral (x4/x2) + (1/2) Integral (1/x2)

And so on..
 
  • #14
Thanks everyone for helping me out!
 

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