# Integral: 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..

## Answers and Replies

tiny-tim
Homework Helper
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 $$\sqrt(1+(\frac{x^4-1}{2x^2})^2)$$ ?

Hi Alexx1! Welcome to PF!

Do you mean $$\sqrt(1+(\frac{x^4-1}{2x^2})^2)$$ ?
Yes, it's that integral ;)

tiny-tim
Homework Helper
Yes, it's that integral ;)
ok, then (without integrating) expand everything inside the √, and then simplify …

what do you get?

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

what do you get?
(8x^8+2x^4+x)/(4x^4)

tiny-tim
Homework Helper
(try using the X2 tag just above the Reply box )
(8x^8+2x^4+x)/(4x^4)
uhhh? not even close

(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
Homework Helper
(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
Homework Helper
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
Homework Helper
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!

(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
Homework Helper
But now I don't know what to do with : √x8+2x4+1
come on, think!

(or put x4 = y )

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..

Thanks everyone for helping me out!