Evaluate the integral: I = integral dx/(cube root
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt[3]{1+x^3}}. $$
Explanation of Method
This is a binomial differential $x^m(1+x^n)^p$ with $m=0,n=3,p=-\tfrac13$. Chebyshev's theorem: $\tfrac{m+1}n=\tfrac13$ is not an integer, $p=-\tfrac13$ is not an integer, but $\tfrac{m+1}n+p=\tfrac13-\tfrac13=0$ is an integer — so the integral is elementary, via the Case 3 substitution $z^3=x^{-3}+1$.
- Verify elementariness via Chebyshev's theorem.
- Substitute $z^3=x^{-3}+1$.
- Reduce to a rational integral in $z$.
- Integrate and back-substitute.
Step 1 · Substitute
Let $z=\dfrac{(1+x^3)^{1/3}}x$, i.e. $z^3=x^{-3}+1$, so $x^3=\dfrac1{z^3-1}$, $x=(z^3-1)^{-1/3}$:
$$ dx = -z^2(z^3-1)^{-4/3}\,dz. $$
Also $1+x^3=\dfrac{z^3}{z^3-1}$, so $(1+x^3)^{-1/3}=\dfrac{(z^3-1)^{1/3}}z$.
Step 2 · Combine
$$ \frac{dx}{\sqrt[3]{1+x^3}} = \frac{(z^3-1)^{1/3}}z\cdot\left[-z^2(z^3-1)^{-4/3}\right]dz = -\frac{z}{z^3-1}\,dz. $$
This is now a rational function of $z$, confirming elementariness directly.
Step 3 · Integrate the rational function
Standard partial-fraction integration of $\dfrac z{z^3-1}$ gives:
$$ I = -\int\frac{z\,dz}{z^3-1} = -\frac13\ln|z-1| + \frac16\ln\left(z^2+z+1\right) - \frac{\sqrt3}3\arctan\left(\frac{2z+1}{\sqrt3}\right) + C. $$
Step 4 · Back-substitute
With $z=\dfrac{(1+x^3)^{1/3}}x$:
Summary Table
| Chebyshev quantity | Value |
|---|---|
| $m,\,n,\,p$ | $0,\ 3,\ -1/3$ |
| $(m+1)/n+p$ | $0$ (integer → elementary, Case 3) |
Additional Notes
- Major correction: the source concluded this integral "does not have a simple elementary closed form" and presented only a hypergeometric series. Chebyshev's theorem shows the opposite: since $\tfrac{m+1}n+p=0$ is an integer, this integral is elementary. The elementary closed form above was obtained via the Case 3 rationalizing substitution and was verified independently.
- The domain is all real $x$ with $1+x^3\neq0$, i.e. $x\neq-1$.
- The result was verified by differentiation at $x=1,2,3$: the derivative matches the integrand $(1+x^3)^{-1/3}$ exactly at all three points.
Result
The solution to the integral is, with $z=\dfrac{\sqrt[3]{1+x^3}}x$: