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

[Alexx1]

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..

 

[tiny-tim]

Welcome to PF!

Hi Alexx1! Welcome to PF!

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 \sqrt(1+(\frac{x^4-1}{2x^2})^2) ?

 

[Alexx1]

Hi Alexx1! Welcome to PF!



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

Yes, it's that integral ;)

 

[tiny-tim]

Yes, it's that integral ;)

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

what do you get? ​

 

[Alexx1]

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

what do you get? ​

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

 

[tiny-tim]

(try using the X² tag just above the Reply box )

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

uhhh? not even close

 

[Alexx1]

(try using the X² 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)

 

[tiny-tim]

(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!

and the square-root of that is … ?​

 

[Mentallic]

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?

 

[HallsofIvy]

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!

 

[Alexx1]

(please use the X² tag just above the Reply box … it makes it much easier to read)ok!

and the square-root of that is … ?​

= √x⁸+2x⁴+1/√4x⁴

= √(x⁴+1)²/2x²

= x⁴+1/2x²

 

[tiny-tim]

But now I don't know what to do with : √x⁸+2x⁴+1

come on, think!

(or put x⁴ = y )​

 

[Alexx1]

come on, think!

(or put x⁴ = y )​

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..

 

[Alexx1]

Thanks everyone for helping me out!