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

  • Thread starter Alexx1
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  • #1
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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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  • #2
tiny-tim
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Welcome to PF!

Hi Alexx1! Welcome to PF! :smile:
sqrt(1+((x^4-1)/(2x^2))^2)
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 [tex]\sqrt(1+(\frac{x^4-1}{2x^2})^2)[/tex] ? :confused:
 
  • #3
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Hi Alexx1! Welcome to PF! :smile:



Do you mean [tex]\sqrt(1+(\frac{x^4-1}{2x^2})^2)[/tex] ? :confused:
Yes, it's that integral ;)
 
  • #4
tiny-tim
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Yes, it's that integral ;)
ok, then (without integrating) expand everything inside the √, and then simplify …

what do you get? :smile:
 
  • #5
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ok, then (without integrating) expand everything inside the √, and then simplify …

what do you get? :smile:
(8x^8+2x^4+x)/(4x^4)
 
  • #6
tiny-tim
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(try using the X2 tag just above the Reply box :wink:)
(8x^8+2x^4+x)/(4x^4)
uhhh? :confused: not even close :redface:
 
  • #7
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(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)
 
  • #8
tiny-tim
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(please use the X2 tag just above the Reply box … it makes it much easier to read)
… (x^8+2x^4+1)/(4x^4)
ok! :smile:

and the square-root of that is … ?​
 
  • #9
Mentallic
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What should've made factorising easy to spot is that you first expanded
[tex](x^4-1)^2=x^8-2x^4+1[/tex]

and after adding [itex]4x^4[/itex] you now have
[tex]x^8+2x^4+1[/tex]

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

That's used a lot to produce easy "arc length" problems!
 
  • #11
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(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
tiny-tim
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But now I don't know what to do with : √x8+2x4+1
come on, think! :smile:

(or put x4 = y :wink:)
 
  • #13
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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
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Thanks everyone for helping me out!
 

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