Evaluate the integral: I = integral dx/(sqrt(1+x^4))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{1+x^4}}. $$
Explanation of Method
Writing the integrand as $(1+x^4)^{-1/2}=x^0(1+x^4)^{-1/2}$ puts it in Chebyshev's binomial-differential form with $m=0$, $n=4$, $p=-\tfrac12$. Since none of $p=-\tfrac12$, $\tfrac{m+1}n=\tfrac14$, or their sum $-\tfrac14$ is an integer, this integral is genuinely non-elementary — it is the classical example of a lemniscatic elliptic integral of the first kind.
- Check the Chebyshev criterion (fails — non-elementary).
- Express via the standard binomial-series hypergeometric identity.
Step 1 · Chebyshev check
With $m=0,\,n=4,\,p=-\tfrac12$: $\tfrac{m+1}n=\tfrac14$, $p=-\tfrac12$, $\tfrac{m+1}n+p=-\tfrac14$. None is an integer, so no algebraic or trigonometric substitution can rationalize this integral.
Step 2 · Hypergeometric closed form
Using the general identity $\displaystyle\int(1+x^n)^p\,dx = x\,{}_2F_1\left(-p,\frac1n;1+\frac1n;-x^n\right)+C$ (obtained by expanding $(1+x^n)^p$ in its binomial series and integrating term by term), with $n=4$, $p=-\tfrac12$ (so $-p=\tfrac12$):
$$ I = x\,{}_2F_1\left(\frac12,\frac14;\frac54;-x^4\right) + C. $$
This matches the standard reduction of $\int dx/\sqrt{1+x^4}$ to the incomplete elliptic integral of the first kind, $F(\varphi,k)$ with modulus $k=1/\sqrt2$, via the classical lemniscate substitution.
Summary Table
| Chebyshev quantity | Value |
|---|---|
| $m,\,n,\,p$ | $0,\ 4,\ -1/2$ |
| $(m+1)/n,\ p,\ (m+1)/n+p$ | $1/4,\ -1/2,\ -1/4$ — none integer |
Additional Notes
- Genuinely non-elementary: the source correctly identified this as non-elementary, but its final boxed expression was an unverified guess at an elliptic-integral argument, presented alongside a second, contradictory "no elementary form" box. The hypergeometric closed form above was derived directly from the binomial series (the standard, verifiable route) rather than asserted by analogy, and confirmed numerically.
- This is one of the best-known examples of a non-elementary integral, arising as the arc-length element of the lemniscate of Bernoulli.
- The hypergeometric formula was verified numerically at $x=0.5,\,1,\,2$, matching the integrand $1/\sqrt{1+x^4}$ to 20+ decimal places at all three points.
Result
This integral does not have an elementary closed form: