Evaluate the integral: I = integral dx/(cube root
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt[3]{(x+1)^2(x-2)^2}} = \int (x+1)^{-2/3}(x-2)^{-2/3}\,dx. $$
Explanation of Method
Here the exponents at the two finite singular points are $p=-\tfrac23$ and $q=-\tfrac23$, and their sum $p+q=-\tfrac43$ is not an integer, nor is either exponent alone. For $\int(x-a)^p(x-b)^q\,dx$, elementariness requires $p$, $q$, or $p+q$ to be an integer — none holds here, so (unlike Problem 328) this integral is genuinely non-elementary. A genus computation on the associated triple cover confirms this: the exponents $-\tfrac23$ at $x=-1,\,2,\,\infty$ (each totally ramified in a degree-3 cover) give genus $1$ by Riemann–Hurwitz. The best closed form is a hypergeometric function.
- Substitute $t=(x+1)/(x-2)$ to isolate the singular structure.
- Confirm non-elementariness via the exponent/genus criterion.
- Express the result via the incomplete Beta / hypergeometric integral.
Step 1 · Substitute
Let $t=\dfrac{x+1}{x-2}$, so $x=\dfrac{2t+1}{t-1}$, $dx=\dfrac{-3}{(t-1)^2}dt$, $x+1=\dfrac{3t}{t-1}$, $x-2=\dfrac3{t-1}$. Substituting:
$$ I = -3^{-1/3}\int t^{-2/3}(t-1)^{-2/3}\,dt. $$
Step 2 · Recognize the exponent obstruction
The exponents at $t=0$ and $t=1$ are both $-\tfrac23$; the exponent at $t=\infty$ (from the large-$t$ behavior $t^{-4/3}$) is also $-\tfrac23$. Since none of $-\tfrac23$, $-\tfrac23$, or their sum $-\tfrac43$ is an integer, the substitution $v=t^{1/3}$ that worked for Problem 328 does not rationalize this integral — it produces a genuine degree-3 algebraic curve of genus $1$, not genus $0$. No further elementary substitution exists.
Step 3 · Hypergeometric closed form
For $0<t<1$, using the standard identity $\displaystyle\int_0^t s^{\alpha-1}(1-s)^{\beta-1}ds=\frac{t^\alpha}\alpha\,{}_2F_1(\alpha,1-\beta;\alpha+1;t)$ with $\alpha=\tfrac13,\ \beta=\tfrac13$ (matching $t^{-2/3}(1-t)^{-2/3}=t^{\alpha-1}(1-t)^{\beta-1}$):
$$ \int t^{-2/3}(1-t)^{-2/3}\,dt = 3t^{1/3}\,{}_2F_1\left(\frac13,\frac23;\frac43;t\right) + C. $$
Multiplying by $-3^{-1/3}$ and using $t=(x+1)/(x-2)$:
Summary Table
| Quantity | Value |
|---|---|
| Exponents at $t=0,1,\infty$ | $-2/3,\ -2/3,\ -2/3$ |
| Genus (Riemann–Hurwitz) | $1$ (non-elementary) |
Additional Notes
- Major rewrite — genuinely non-elementary: the source's Step 5 gave up on a rigorous derivation ("this integral is still complex... after careful evaluation using Beta function properties") and asserted the boxed answer $-\tfrac{3\sqrt[3]9}2\sqrt[3]{(x+1)/(x-2)}$, which fails numerically (derivative $\approx1.238$ vs. true integrand $\approx0.397$ at $x=3$). Unlike that false claim of a clean elementary answer, the exponent/genus analysis above shows this integral genuinely has no elementary closed form; the hypergeometric expression is the best available.
- Contrast with Problem 328, which has the same shape of integrand but exponent sum $-1$ (an integer) — that one is elementary. The difference in exponents is what separates the two cases.
- The hypergeometric formula above and its derivative were verified numerically at $x=-2,-3,-5$ (the branch where $t=(x+1)/(x-2)\in(0,1)$), matching the integrand $(x+1)^{-2/3}(x-2)^{-2/3}$ exactly at all three points.
Result
This integral does not have an elementary closed form. Writing $t=\dfrac{x+1}{x-2}$: