Indefinite Integration

[jacks]

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

[Papapetros Vaggelis]

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