- #1

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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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- Thread starter Alexx1
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- #1

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Can someone explain me how to solve an integral like this?

I tried it, but I have absolutely no idea how to solve it..

- #2

tiny-tim

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Hi Alexx1! Welcome to PF!

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

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] ?

- #3

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Hi Alexx1! Welcome to PF!

Do you mean [tex]\sqrt(1+(\frac{x^4-1}{2x^2})^2)[/tex] ?

Yes, it's that integral ;)

- #4

tiny-tim

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Yes, it's that integral ;)

ok, then (

what do you get?

- #5

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

what do you get?

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

- #6

tiny-tim

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(try using the X^{2} tag just above the Reply box )

uhhh?*not even close*

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

uhhh?

- #7

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(try using the X^{2}tag just above the Reply box )

uhhh?not even close

((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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… (x^8+2x^4+1)/(4x^4)

ok!

and the *square-root* of that is … ?

- #9

Mentallic

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[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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That's used a lot to produce easy "arc length" problems!

- #11

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(pleaseuse the X^{2}tag just above the Reply box … it makes itmucheasier to read)

ok!

and thesquare-rootof that is … ?

= √x

= √(x

= x

- #12

tiny-tim

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But now I don't know what to do with : √x^{8}+2x^{4}+1

come on,

(or put x^{4} = y )

- #13

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come on,think!

(or put x^{4}= y )

Ok, I found it!

= √x

= √(x

= x

==> Integral x

= (1/2) Integral (x

And so on..

- #14

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Thanks everyone for helping me out!

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