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What substitution is useful for this integral?

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

Substitutions $u=x^{\frac{2}{3}},u=(x+1)^{\frac{2}{3}},u=(x-2)^{\frac{2}{3}}$ are not working.

Can't find useful trigonometric substitution.

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  • $\begingroup$ What class is this for? Wolfram evaluated the integral for me and it involves a hypergeometric function, so some context on where this problem is coming from would be nice. $\endgroup$ – TomGrubb Jan 3 '16 at 23:18
  • $\begingroup$ @bburGsamohT Calculus 2 class, exam paper problem. $\endgroup$ – user300048 Jan 3 '16 at 23:22
  • $\begingroup$ When you ask for a sub, do you need a sub that leads to an elementary solution or simply one that puts the integral into a "nicer form"? $\endgroup$ – Brevan Ellefsen Jan 3 '16 at 23:26
  • $\begingroup$ @Brevan Ellefsen I don't think that any substitution could immediately lead to solution, but if you have an idea on that (or to simplify an integral) then suggest it. $\endgroup$ – user300048 Jan 3 '16 at 23:30
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Change variable to $$x - \frac12 = \frac32 u \iff u = \frac{2x-1}{3} \quad\text{ followed by }\quad (u^2-1)^{1/3} = v \iff u = \sqrt{1+v^3}$$ We have $$\int\frac{dx}{\sqrt[3]{(x+1)^2(x-2)^2}} = \int\frac{dx}{(x^2-x-2)^{2/3}} = \int\frac{dx}{\left(\left(x-\frac12\right)^2-\frac{9}{4}\right)^{2/3}}\\ = \left(\frac{2}{3}\right)^{1/3}\int\frac{du}{(u^2-1)^{2/3}} = \left(\frac{2}{3}\right)^{1/3}\int\frac{d\sqrt{1+v^3}}{v^2} = \left(\frac{3}{2}\right)^{2/3}\int \frac{dv}{\sqrt{1+v^3}} $$ Using result from this answer, the integral becomes

$$\frac{3^{5/12}}{2^{2/3}} \left.F\left(\cos^{-1}\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\sqrt[3]{\frac49(x+1)(x-2)}}-1\right)\right|\frac{2+\sqrt{3}}{4}\right) + \text{constant}$$ where $\displaystyle\;F(\phi|m) = \int_0^\phi \frac{d\theta}{\sqrt{1-m(\sin\theta)^2}}\;$ is the incomplete elliptic integral of the first kind.

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  • $\begingroup$ Nice answer indeed ! $\endgroup$ – Claude Leibovici Jan 4 '16 at 10:18
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I found an interesting substitution by a bit of trial and error. $$\begin{align} x&=\frac{1+3\sin(u)}{2}\\ \therefore dx&=\frac{3\cos(u)}{2}du\\ x+1&=\frac{3+3\sin(u)}{2}&&=\frac{3(1+\sin(u))}{2}\\ x-2&=\frac{3\sin(u)-3}{2}&&=\frac{3(\sin(u)-1)}{2}\\ \therefore (x+1)(x-2)&=\frac{9(\sin^2(u)-1)}{4}&&=-\frac{9\cos^2(u)}{4} \end{align}$$ Using these leads to this interesting final integral:$$\left(\frac{2}{3}\right)^{\frac{1}{3}}\int\frac{du}{\cos^{\frac{1}{3}}(u)}$$

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  • $\begingroup$ Unfortunately this integral doesn't seem any easier to solve $\endgroup$ – Mufasa Jan 4 '16 at 0:05
  • $\begingroup$ I'm pretty sure that this integral is simple enough to show though that no elementary solution for the antiderivative exists. Perhaps the hypergeometric could be represented as a simpler Special Function, but I sincerely doubt this will turn out to be elementary $\endgroup$ – Brevan Ellefsen Jan 4 '16 at 1:03
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As commented by Brevan Ellefsen, there is no elementary solution for $$I=\left(\frac{2}{3}\right)^{\frac{1}{3}}\int\frac{du}{\cos^{\frac{1}{3}}(u)}$$ but there is effectively a solution using hypergeometric function. It seems to be $$I=-\left(\frac{3}{2}\right)^{2/3} \cos ^{\frac{2}{3}}(u) \, _2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(u)\right)$$ where appears the Gaussian (sometimes called ordinary) hypergeometric function.

In terms of $x$, the antiderivative is then $$I=\int \frac{1}{\sqrt[3]{(x+1)^2(x-2)^2}}\, dx= \sqrt[3]{3(x-2)} \,\,\, _2F_1\left(\frac{1}{3},\frac{2}{3};\frac{4}{3};\frac{2-x}{3}\right)$$ Back to definitions (and getting rid of Pochhammer symbols), $$I=\sum_{n=0}^\infty \frac{3^{-4 n} \,\Gamma \left(\frac{1}{3}\right)\, \Gamma (3 n)}{n!\, \Gamma (n)\, \Gamma \left(n+\frac{4}{3}\right)} (2-x)^n$$

Edit

The coefficients in the last expression vary extremely fast ( the first is $1$, the tenth is $\approx 1.851\times 10^{-7}$, the hundredth is $\approx 1.025\times 10^{-51}$). For large values of $n$, they are $$\log(c_n)=-n \log (3)-\frac{4}{3} \log (n)+\log \left(\frac{\Gamma \left(\frac{1}{3}\right)}{2 \sqrt{3} \pi }\right)-\frac{4}{9 n}+O\left(\frac{1}{n^2}\right)$$

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What substitution is useful for this integral ?


None. The integrand does not possess any elementary anti-derivative. See Liouville's theorem

and the Risch algorithm for more information. However, all definite integrals of the following

form: $\displaystyle\int_a^b\Big[(x-x_1)(x-x_2)\Big]^r~dx,$ with $a,b\in\{x_1,x_2,\pm\infty\},$ can be evaluated in terms of the

beta and $\Gamma$ functions, assuming, of course, that they converge in the first place. This should

come as no surprise, given the fact that Mufasa has already been able to rewrite the original

expression as a Wallis integral, whose relation to the aforementioned special functions is well

known. Not to mention the fact that Claude Leibovici's hypergeometric series can also be

expressed in terms of the incomplete beta function. Thus, for $x_{1,2}=\{-1,2\}$ and $r=-\dfrac23,$

we have the following result:

$$\begin{align} \int_{-\infty}^\infty\Big[(x+1)(x-2)\Big]^{-\tfrac23}~dx ~&=~3\int_{-\infty}^{-1}\Big[(x+1)(x-2)\Big]^{-\tfrac23}~dx~=~ \\\\ ~&=~3\int_{-1}^2\Big[(x+1)(x-2)\Big]^{-\tfrac23}~dx~=~ \\\\ ~&=~3\int_2^\infty\Big[(x+1)(x-2)\Big]^{-\tfrac23}~dx~=~ \\\\ ~&=~3\cdot\frac{B\Big(\tfrac13~,~\tfrac16\Big)}{\sqrt[3]{12}} \end{align}$$

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  • 1
    $\begingroup$ Nice answer for sure ! Cheers. $\endgroup$ – Claude Leibovici Jan 4 '16 at 10:22

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