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

• Alexx1
In summary, the integral of sqrt(1+((x^4-1)/(2x^2))^2) can be simplified using the substitution method. By letting u=(x^4-1)/(2x^2), the integral can be rewritten as sqrt(1+u^2) du, which can then be solved by using trigonometric substitution. This method allows for the integration of complex expressions involving square roots and polynomials.
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..

Welcome to PF!

Hi Alexx1! Welcome to PF!
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 $$\sqrt(1+(\frac{x^4-1}{2x^2})^2)$$ ?

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

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

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?

tiny-tim said:
ok, then (without integrating) expand everything inside the √, and then simplify …

what do you get?

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

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

uhhh? not even close

tiny-tim said:
(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)

(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!

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?

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!

tiny-tim said:
(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

Alexx1 said:
But now I don't know what to do with : √x8+2x4+1

come on, think!

(or put x4 = y )

tiny-tim said:
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!

## 1. What is the integral of sqrt(1+((x^4-1)/(2x^2))^2)?

The integral of sqrt(1+((x^4-1)/(2x^2))^2) is a complex integral that cannot be expressed in terms of elementary functions. It can be approximated using numerical methods or solved using specialized techniques such as elliptic integrals.

## 2. How do you solve the integral sqrt(1+((x^4-1)/(2x^2))^2)?

The integral sqrt(1+((x^4-1)/(2x^2))^2) can be solved using specialized techniques such as elliptic integrals or by approximating it using numerical methods. It cannot be solved using traditional methods of integration.

## 3. What is the purpose of the sqrt(1+((x^4-1)/(2x^2))^2) integral?

The integral sqrt(1+((x^4-1)/(2x^2))^2) is often used in physics and engineering to calculate the arc length of a curve. It can also be used in other applications such as calculating the surface area of certain shapes.

## 4. Can the integral sqrt(1+((x^4-1)/(2x^2))^2) be simplified?

No, the integral sqrt(1+((x^4-1)/(2x^2))^2) cannot be simplified using traditional methods of integration. It can only be approximated using numerical methods or solved using specialized techniques.

## 5. Is there a way to express the integral sqrt(1+((x^4-1)/(2x^2))^2) in terms of known functions?

No, the integral sqrt(1+((x^4-1)/(2x^2))^2) cannot be expressed in terms of elementary functions such as polynomials, trigonometric functions, or exponential functions. It can only be approximated using numerical methods or solved using specialized techniques.

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