Integrate $\int\frac{x}{\sqrt[3]{x^{3}-3x-2}}dx$

Question

I need help in Integrating $$\int\frac{x}{\sqrt[3]{x^{3}-3x-2}}dx$$

This question came in the MIT-BEE competition in the year $2025$

My Approach

First of all I tried to solve this integral by factorizing the denominator.

Clearly $(x+1)$ is a factor of the polynomial $(x^3-3x-2)$

Therefore I tried to evaluate the integral in this way :-

Let

$$I= \int \frac{x}{\sqrt[3]{x^{3}-3x-2}}dx$$

$$\implies I=\int\frac{x}{\sqrt[3]{(x+1)^{2}(x-2)}}dx$$

$$\implies I=\int\frac{x}{(x+1)^{\frac{2}{3}}(x-2)^{\frac{1}{3}}}dx$$

Now I thought of substituting $x=\tan^{2}\theta \implies dx=2(\tan\theta)(\sec^{2}\theta)d\theta$

$$\implies I =2\int\frac{(\tan^{3}\theta)(\sec^{2}\theta)}{(\sec^{\frac{4}{3}}\theta)(\tan^{2}\theta-2)^{\frac{1}{3}}}d\theta$$

Now I can't understand how to proceed further from the last step

Please help me.

Integral calculator gives the answer $(x-2)^{\frac{2}{3}}\sqrt[3]{x+1}+C$, you may check the steps if you are interested in other methods. — Bowei Tang, Jun 27 '25 at 10:26
Not quite a duplicate, but similar: https://math.stackexchange.com/questions/1628155/the-integral-int-frac-rm-dxx3-frac87x-2-frac-67 — Travis Willse, Jun 27 '25 at 11:49
Does anything nice happen if you rewrite $x$ as $\frac13[2(x+1)+(x-2)]$? I may have to try later. — Mike, Jun 27 '25 at 12:48

score 5 · Answer 1 · answered Jun 27 '25 at 11:09

It's not as nice as the ansatz computation, but the factorization of the argument of the square root suggests a substitution like $$u = \sqrt[3]{\frac{x + 1}{x - 2}}$$ (or its reciprocal), which rationalizes the integrand, yielding $$-3 \int \frac{(1 + 2 u^3) \,du}{(u^3 - 1)^2} = \frac{3 u}{u^3 - 1} + C = (x + 1)^\frac13 (x - 2)^\frac23 + C .$$

score 4 · Answer 2 · answered Jun 27 '25 at 20:01

All right, giving my comment a try. We'll take it from the factored version.

$$\int\dfrac x{\sqrt[3]{(x+1)^2(x-2)}}dx=\frac13\int\dfrac{2(x+1)+(x-2)}{\sqrt[3]{(x+1)^2(x-2)}}dx=$$ $$\frac23\int\dfrac{x+1}{\sqrt[3]{(x+1)^2(x-2)}}dx+\frac13\int\dfrac{x-2}{\sqrt[3]{(x+1)^2(x-2)}}dx=$$ $$\frac23\int(x+1)^\frac13(x-2)^{-\frac13}dx+\frac13\int(x+1)^{-\frac23}(x-2)^\frac23$$ This looks like the product rule from here. $\frac{d}{dx}(x-2)^\frac23=\frac23(x-2)^{-\frac13}$ and $\frac{d}{dx}(x+1)^\frac13=\frac13(x+1)^{-\frac23}$. So our last integral is equal to $$(x+1)^\frac13(x-2)^\frac23+C=\sqrt[3]{(x+1)(x^2-4x+4)}+C=\sqrt[3]{x^3-3x^2+4}+C$$

I will also give it a try by your Process @Mike – Dev Jun 28 '25 at 11:43 — Dev, Jun 28 '25 at 11:43

Travis Willse · Answer 3 · 2025-06-27T14:09:22.323

The form $\int x (x - a)^p (x - b)^q \,dx$ suggests trying the ansatz

$$(x - a)^r (x - b)^s + C .$$

Differentiating gives $$\int [(r + s) x - (a s + b r)] (x - a)^{r - 1} (x - b)^{s - 1} = (x - a)^r (x - b)^s + C .$$ Setting $a = -1, b = 2$ and comparing coefficients gives $r = \frac13, s = \frac23$. In retrospect, we can see that the ansatz only works because the constants in the original integral satisfy $a(q + 1) + b (p + 1) = 0$.

Integrate $\int\frac{x}{\sqrt[3]{x^{3}-3x-2}}dx$

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