For major players only: Integrate ∫ 1/√(1+x^4) dx

I broke this down using trig substitution (triangle: 1, x^2, √(x^4+1) to

∫ 1/√(2sin2ø) dø

If you could integrate in terms of ø or x it would be amazing.

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- Dec 11th 2008, 12:03 PMcdrappi2552For major players only: Integrate ∫ 1/√(1+x^4) dx...?
For major players only: Integrate ∫ 1/√(1+x^4) dx

I broke this down using trig substitution (triangle: 1, x^2, √(x^4+1) to

∫ 1/√(2sin2ø) dø

If you could integrate in terms of ø or x it would be amazing. - Dec 11th 2008, 01:11 PMLaurent
The answer is a special function, called the "inverse hyperbolic lemniscatic sine"... The inverse lemniscatic sine equals $\displaystyle \int_0^x \frac{dt}{\sqrt{1-t^4}}$ (a generalization of the arcsinus, relating to the lemniscate curve like the sinus relates to the circle...), and its hyperbolic counterpart is the function you need: $\displaystyle \int_0^x\frac{dt}{\sqrt{1+t^4}}$. It is tightly connected to elliptic functions.

In a word: there is no expression for this anti-derivative in terms of "elementary functions"...