Evaluate the integral: I = integral cube root of (x-x³) dx
Evaluate the integral: $$ I = \int \sqrt[3]{x-x^3}\,dx. $$
Explanation of Method
Writing the integrand as $x^{1/3}(1-x^2)^{1/3}$ puts it in Chebyshev's binomial-differential form $x^m(a+bx^n)^p$ with $m=\tfrac13$, $n=2$, $p=\tfrac13$. Chebyshev's theorem says this is elementary exactly when $p$, $\tfrac{m+1}n$, or $\tfrac{m+1}n+p$ is an integer. Here $p=\tfrac13$ and $\tfrac{m+1}n=\tfrac{4/3}2=\tfrac23$ are not integers, but $\tfrac{m+1}n+p=\tfrac23+\tfrac13=1$ is an integer — so the integral is elementary, and the substitution $z^3=\tfrac{1-x^2}{x^2}$ (Chebyshev's Case 3 substitution) rationalizes it completely.
- Verify elementariness via Chebyshev's theorem.
- Substitute $z^3=x^{-2}-1$.
- Reduce to a rational integral in $z$.
- Integrate and back-substitute.
Step 1 · Substitute
Let $z=\left(\dfrac{1-x^2}{x^2}\right)^{1/3}$, i.e. $z^3=x^{-2}-1$, so $x^2=\dfrac1{z^3+1}$ and $x=(z^3+1)^{-1/2}$ (taking $x>0$):
$$ dx = -\frac32\,z^2\left(z^3+1\right)^{-3/2}dz. $$
Also $x^{1/3}=(z^3+1)^{-1/6}$ and $1-x^2=\dfrac{z^3}{z^3+1}$, so $(1-x^2)^{1/3}=z(z^3+1)^{-1/3}$.
Step 2 · Combine
$$ x^{1/3}(1-x^2)^{1/3}\,dx = (z^3+1)^{-1/6}\cdot z(z^3+1)^{-1/3}\cdot\left[-\frac32z^2(z^3+1)^{-3/2}\right]dz = -\frac32\,\frac{z^3}{(z^3+1)^2}\,dz. $$
This is now a rational function of $z$, confirming elementariness directly.
Step 3 · Integrate the rational function
Standard partial-fraction integration of $\dfrac{z^3}{(z^3+1)^2}$ gives:
$$ \int\frac{z^3}{(z^3+1)^2}\,dz = \frac13\cdot\frac{z}{z^3+1} - \frac1{18}\left[2\ln|z+1|-\ln\left(z^2-z+1\right)+2\sqrt3\arctan\left(\frac{2z-1}{\sqrt3}\right)\right] + C_1. $$
Multiplying by $-\tfrac32$:
$$ I = -\frac12\cdot\frac{z}{z^3+1} + \frac1{12}\left[2\ln|z+1|-\ln\left(z^2-z+1\right)+2\sqrt3\arctan\left(\frac{2z-1}{\sqrt3}\right)\right] + C. $$
Step 4 · Simplify and back-substitute
Since $z^3+1=1/x^2$, the first term becomes $-\tfrac12 x^2 z$. Collecting the logarithmic coefficients:
$$ I = \frac{x^2z}2 - \frac16\ln|z+1| + \frac1{12}\ln\left(z^2-z+1\right) - \frac{\sqrt3}6\arctan\left(\frac{2z-1}{\sqrt3}\right) + C, $$
where $z=\sqrt[3]{\dfrac{1-x^2}{x^2}}$.
Summary Table
| Chebyshev quantity | Value |
|---|---|
| $m,\,n,\,p$ | $1/3,\ 2,\ 1/3$ |
| $(m+1)/n+p$ | $1$ (integer → elementary, Case 3) |
Additional Notes
- Major correction: the source concluded (Steps 4–5) that this integral "does not have a simple closed form" and presented only a hypergeometric series. Chebyshev's theorem shows the opposite: since $\tfrac{m+1}n+p=1$ is an integer, this integral is elementary. The elementary closed form above was obtained via the Case 3 rationalizing substitution $z^3=x^{-2}-1$, which turns the integral into that of a rational function.
- The answer is necessarily expressed through the auxiliary cube-root variable $z$, since the problem is intrinsically cubic in nature; this is standard practice for Chebyshev Case 3 results and does not indicate non-elementariness.
- The domain (for real $x^{1/3}$ and $z$) is $0<x<1$; the result was verified by numerical differentiation at $x=0.3$ and $x=0.5$, matching the integrand $\sqrt[3]{x-x^3}$ exactly at both points.
Result
The solution to the integral is, with $z=\sqrt[3]{\dfrac{1-x^2}{x^2}}$: