# Integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx

• DJ-Math
In summary, the conversation discusses how to solve the integral ∫(x2-1)/(1+x2)*1/(1+x4)(1/2)dx using the substitution u2 = x2 + 1/x2. The speaker has attempted to use trig substitution and the u substitution, but is unsure if their solutions are correct. They also mention that this question has been previously discussed on the platform. After some discussion, it is suggested to use an inverse tangent function in the solution, specifically (1/sqrt2)*arctan(sqrt((x^2 + 1/x^2)/2)).
DJ-Math
Member warned about posting with no template and no effort shown
Hi,
I'm trying to solve ∫(x2-1)/(1+x2)*1/(1+x4)(1/2)dx

I'm apparently meant to get some non-complex result, the question suggests to use the substitution u2 = x2 + 1/x2
But I haven't gotten anywhere with this.

Any methods or suggestions (or the solution) would be much appreciated!
Thanks

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If you take the hint, ##u^2 = x^2 + \frac{1}{x^2} ## then ## du = \frac{x - \frac {1}{x^3}}{\sqrt{x^2 + \frac{1}{x^2}}}dx= \frac{x^2 - \frac {1}{x^2}}{\sqrt{x^4 +1}}dx##
So try to substitute in the du and try simplifying from there.

DJ-Math said:
Hi,
I'm trying to solve ∫(x2-1)/(1+x2)*1/(1+x4)(1/2)dx

I'm apparently meant to get some non-complex result, the question suggests to use the substitution u2 = x2 + 1/x2
But I haven't gotten anywhere with this.

Any methods or suggestions (or the solution) would be much appreciated!
Thanks

This question was posed here about 2-3 months ago, and was discussed thoroughly (and solved) then. If you look for it you will find it.

Here's my two attempts at a solution (photo attached - second photo is a little messy) - I tried a trig substitution and the u substitution and a following simplification.

I also looked through the past 5 months of "integration" related questions and couldn't find this particular question so if you had a direct link to it that would be much appreciated.

Thanks

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I THINK I GOT IT. If someone could check the photo for errors that'd be great.

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Wasn't x^2+1/x^2= u^2?
I think you may need an inverse tangent function in that case.

RUber said:
Wasn't x^2+1/x^2= u^2?
I think you may need an inverse tangent function in that case.

Right you are! I'll fix that up.

Should be:
(1/sqrt2)*arctan(sqrt((x^2 + 1/x^2)/2)) yeah?

Last edited:
That is what WolframAlpha.com says.

DJ-Math

## 1. What is the formula for finding the integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx?

The formula for finding the integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx is ∫(1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx = tan^-1(x) + C.

## 2. How do you solve for the indefinite integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx?

To solve for the indefinite integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx, use the substitution method by letting u = x^2 + 1. Then, the integral becomes ∫(1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx = (1/2)∫1/u^(1/2)du. This can be solved using the power rule and the resulting answer can be rewritten in terms of x.

## 3. Is it possible to find the definite integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx?

Yes, it is possible to find the definite integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx. However, the resulting answer will involve the use of special functions such as the incomplete elliptic integral of the first kind or the Gauss hypergeometric function.

## 4. What is the substitution used to simplify the integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx?

The substitution used to simplify the integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx is u = x^2 + 1. This substitution allows us to use the power rule to find the integral.

## 5. Can the integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx be solved without using substitution?

No, it is not possible to solve the integral of (1-x^2)/(1+x^2)*(1/(1+x^4)^(1/2))dx without using substitution. Other methods such as integration by parts or partial fractions will not lead to a simpler solution. Therefore, substitution is the most efficient method for solving this integral.

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