Matematické Fórum

stuart clark · 08. 08. 2013 10:42

[1] $\int \sqrt[3]{\frac{x}{1-x}}dx$

[2] $\int \sqrt[3]{\frac{1-x}{x}}dx$

vanok · 08. 08. 2013 19:23 — Editoval vanok (08. 08. 2013 19:24)

Hi,
It is a question of Abel's integral.
In fact one know that
$\int R(x; \sqrt[n]{\frac {ax+b}{cx+d}})dx$
with $ad-cd \ne 0$
$R$ rational fraction of two variables

Used substitution
$t= \sqrt[n]{\frac {ax+b}{cx+d}}$
so
$x= g(t)=\frac {dt^n-b}{a-ct^n}$ and $dx=g'(t)dt$

This transforms the initial problem into calculation of one integral of one rational fraction.

Brano · 08. 08. 2013 21:59

try
http://um.mendelu.cz/maw-html/index.php … m=integral
it provides step-by-step calculation

stuart clark

Thanks ↑ Brano: got it

Let $\displaystyle{I=\int \sqrt[3]{\frac{x}{1-x}}\,dx}$

By the substitution $\displaystyle{\frac{x}{1-x}=t^3}$ , we have

$\displaystyle{x=\frac{t^3}{t^3+1}}$ and $\displaystyle{dx=\frac{3\,t^2}{\left(t^3+1\right)^2}\,dt}$ .

So,

$\displaystyle{\begin{aligned} I&=\int t\,\frac{3\,t^2}{\left(t^3+1\right)^2}\,dt\\&=\int t\,d\left(\frac{-1}{t^3+1}\right)\\&=-\frac{t}{t^3+1}+\int \frac{1}{t^3+1}\,dt\end{aligned}}$

It is easy to prove that $\displaystyle{\frac{1}{t^3+1}=\frac{1}{3\left(t+1\right)}+\frac{2-t}{3\left(t^2-t+1\right)}}$

$\displaystyle{\int \frac{1}{3\left(t+1\right)}\,dt=\frac{1}{3}\,\ln\left|t+1\right|+c\,\,,c\in\mathbb{R}}$

$\displaystyle{\begin{aligned} \int \frac{2-t}{t^2-t+1}\,dt&=-\frac{1}{2}\int \frac{\left(2\,t-1\right)-3}{t^2-t+1}\,dt\\&=-\frac{1}{2}\int \frac{2\,t-1}{t^2-t+1}\,dt+\frac{3}{2}\int \frac{1}{t^2-t+1}\,dt\\&=-\frac{1}{2}\,\ln \left(t^2-t+1\right)+\frac{3}{2}\int \frac{4}{3+\left(2\,t-1\right)^2}\,dt\\&=-\frac{1}{2}\,\ln \left(t^2-t+1\right)+2\int \frac{1}{1+\left(\frac{2\,t-1}{\sqrt{3}}\right)^2}\,dt\\&=-\frac{1}{2}\,\ln\left(t^2-t+1\right)+\sqrt{3}\,\arctan\left(\frac{2\,t-1}{\sqrt{3}}\right)+c \end{aligned}}$

Therefore,

$\displaystyle{I=\int \sqrt[3]{\frac{x}{1-x}}\,dx=-\frac{t}{t^3+1}+\frac{\ln\left|1+t\right|}{3}-\frac{1}{6}\,\ln\left(t^2-t+1\right)+\frac{1}{\sqrt{3}}\,\arctan\left(\frac{2\,t-1}{\sqrt{3}}\right)+c\,\,,c\in\mathbb{R}}$

where $\displaystyle{t=\sqrt[3]{\frac{x}{1-x}}\,,x\neq 1}$

For the second integral, we use the substitution $\displaystyle{x=1-t}$ .

So, $\displaystyle{\int \sqrt[3]{\frac{1-x}{x}}\,dx=-\int \sqrt[3]{\frac{t}{1-t}}\,dt}$