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

[DJ-Math]

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

 

[RUber]

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.

 

[Ray Vickson]

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.

 

[DJ-Math]

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

 

[DJ-Math]

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

 

[RUber]

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

 

[DJ-Math]

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?

 

[RUber]

That is what WolframAlpha.com says.