Evaluating a rational function integral in a quick way

Question

In an recent test I was asked to evaluate the integral $$ \int_0^1 \frac{\sqrt[3]{x^2(1-x)}}{(1+x)^3} \text {d}x $$ in 8 minutes, but I didn't have a clue what to do with it.
After the test, I tried the following approach: substituting $x=\frac{1}{t}, u=\sqrt[3]{t-1}$ gives $$ \int_0^1 \frac{\sqrt[3]{x^2(1-x)}}{(1+x)^3} \text {d}x =\int_1^\infty \frac{\sqrt[3]{t-1}}{(t+1)^3}\text {d}t =\int_0^\infty \frac{3u^3}{(u^3+2)^3}\text {d}u $$ then it becomes a problem of evaluating a rational function integral, but I still don't think it could be done in the time limit.

Is there any quick method to evaluate this? Thanks for any help.

(Note: 1. The integral is $\frac{\pi}{9\times 2^{2/3}\sqrt3}$ according to Wolfram Alpha.
2. It is best to solve it with first-year calculus, and I will understand more easily. Other solutions are also welcomed:)

Answer 1

Substitute $x=\frac{1-t}{1+t}$

\begin{align} &\int_0^1 \frac{\sqrt[3]{x^2(1-x)}}{(1+x)^3} {d} =\frac1{2^{5/3}}\int_0^1 (1-t)^{\frac23}t^{\frac13}dt\\ = & \ \frac{\Gamma(\frac43)\Gamma(\frac53)}{2^{8/3}} = \frac{\Gamma(\frac13)\Gamma(\frac23)}{18\cdot 2^{2/3}} = \frac{\pi}{9\sqrt3\cdot 2^{2/3}} \end{align}

where the reflection formula $\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin(\pi a)}$ is used in the last step.

Answer 2

I would like to start with the following intergral, as you have written: $$\int_0^\infty \frac{3u^3}{(u^3+2)^3}\text {d}u $$ Substitute $u = 2^{\frac{1}{3}}v$ , and the intergral becomes: $$ 2^{\frac{1}{3}}\times \frac{3}{4} \int_0^\infty \frac{v^3}{(v^3+1)^3}\text {d}v $$ Let

$$I_n=\displaystyle\int_0^\infty \frac{1}{(x^3+1)^n}\text {d}x =-\int_0^\infty x\text{d}\frac{1}{(x^3+1)^n}=3n\int_0^\infty \frac{x^3}{(x^3+1)^{n+1}}\text {d}x =3n\int_0^{\infty} \frac{x^3+1-1}{(x^3+1)^{n+1}}\text {d}x =3nI_n-3nI_{n+1}\implies I_{n+1}=\frac{3n-1} {3n}I_n. $$

As

$$I_1=\displaystyle \int_0^\infty \frac{1}{x^3+1}\text {d}x =\frac{2\pi}{3\sqrt{3}},$$

we can easily know that

$$I_2-I_3=\displaystyle\int_0^\infty \frac{v^3}{(v^3+1)^3}\text {d}v =\frac{2\pi}{27\sqrt{3}}.$$

Therefore,

$$\int_0^1 \frac{\sqrt[3]{x^2(1-x)}}{(1+x)^3} \text {d}x =2^{\frac{1}{3}}\times \frac{3}{4} \int_0^\infty \frac{v^3}{(v^3+1)^3}\text {d}v =2^{\frac{1}{3}}\times \frac{3}{4} \times \frac{2\pi}{27\sqrt{3}}=\frac{\pi}{9\times 2^{2/3}\sqrt{3}}$$

as desired.

Answer 3

Here is a way without Gamma functions, and with one well-known integral only.

Starting as @tys:

From OP $$I = \int_0^\infty \frac{3u^3}{(u^3+2)^3}\text {d}u $$ Substituting $u = 2^{\frac{1}{3}}v$ , the integral $I$ becomes: $$ I = 2^{\frac{1}{3}}\times \frac{3}{4} \int_0^\infty \frac{v^3}{(v^3+1)^3}\text {d}v $$

What comes now is a method of parametric integration:

Consider, with $a>0$ and $b>0$ , the more general integral: $$ K(a,b) = \int_0^\infty \frac{1}{a^3v^3+b^3}\text {d}v $$

Notice that $$ \frac{\partial^2 K(a,b)}{\partial a \partial b} = 18 a^2 b^2 \int_0^\infty \frac{v^3}{(a^3v^3+b^3)^3}\text {d}v $$ hence $$ I = 2^{\frac{1}{3}}\times \frac{3}{4\times 18} \lim_{a =1; b = 1}\frac{\partial^2 K(a,b)}{\partial a \partial b} $$

Now the integration in $K(a,b)$ seems complicated, but can be made astonishingly simple. All that is needed is the well-known integral $$ \int\frac{1}{x^2 + 1 }\text {d}x = \tan^{-1}(x)$$

First, let $x = \frac a b v$ to obtain: $$ K(a,b) = \int_0^\infty \frac{1}{a^3v^3+b^3}\text {d}v = \frac1{ a b^2} \int_0^\infty \frac{1}{x^3+1}\text {d}x $$ Then, let $t = 1/x$ which gives: $$ K(a,b) = \frac1{ a b^2} \int_0^\infty \frac{t}{t^3+1}\text {d}t $$ Adding the two formulations of $K(a,b)$ $$ 2 K(a,b) = \frac1{ a b^2} \int_0^\infty \frac{t+1}{t^3+1}\text {d}t = \frac1{ a b^2} \int_0^\infty \frac{1}{(t-\frac 1 2)^2 + \frac 3 4 }\text {d}t = \frac{2}{a b^2\sqrt{3}} \tan^{-1}\frac{2 t - 1}{\sqrt{3}} |_{0}^{\infty}$$ Inserting the limits we have $$ K(a,b) = \frac{1}{a b^2\sqrt{3}} (\frac{\pi} {2}- \frac{-\pi}{6}) = \frac{2 \pi}{3 a b^2\sqrt{3}} $$

Then $$ \frac{\partial^2 K(a,b)}{\partial a \partial b} = \frac{4 π}{3 \sqrt{3} a^2 b^3} $$ and finally

$$ I = 2^{\frac{1}{3}}\times \frac{1}{24} \lim_{a =1; b = 1}\frac{\partial^2 K(a,b)}{\partial a \partial b} = 2^{\frac{1}{3}}\times \frac{1}{24} \times \frac{4 π }{3 \sqrt{3} } =\frac{\pi}{9\times 2^{2/3}\sqrt{3}}$$

as desired.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int_{0}^{1}{\root[\large 3]{\rule{0pt} {3.85mm}x ^{2}\pars{1 - x}} \over \pars{1 + x}^{3}}\dd x} \\[5mm] = & \ {1 \over 2^{5/3}}\bracks{\color{red}{1}^{4/3}\ \pars{1 + \color{red}{1}}^{\color{red}{5/3}}\int_{0}^{1}{x^{\color{red}{5/3}\ -\ 1}\,\, \pars{1 - x}^{\color{red}{4/3}\ -\ 1}\ \over \pars{x + \color{red}{1}}^{\color{red}{5/3}\ +\ \color{red}{4/3}}}\dd x} \\[5mm] = & \ \bbx{\color{#44f}{{1 \over 2^{5/3}}\ \on{B}\pars{{5 \over 3},{4 \over 3}}}} \approx 0.1270\\ & \end{align} where I used G & R $\ds{{\bf 8.380}.7}$ identity.

Answer 5

An elementary way to compute the "crux integral" in the solution by @tys: $$\int_0^\infty \frac{1}{1+x^3}dx = \int_0^1 \frac{1}{1+x^3}dx + \underbrace{\int_1^\infty \frac{1}{1+x^3}dx}_{x\rightarrow 1/x} = \int_0^1 \frac{dx}{x^2-x+1} = \ldots$$