# Integrals of Gradshteyn and Ryzhik: 3.267

We use the following definitions of the gamma and beta functions:

\Gamma(z) = \int\limits_{0}^{\infty} \mathrm{e}^{-t} t^{z-1} dt, \,\, \mathfrak{R}(z) \gt 0

\mathrm{B}(a,b) = \int\limits_{0}^{1} t^{a-1} (1-t)^{b-1} dt
= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, \,\, \mathfrak{R}(a) \gt 0 \,\, \mathrm{and} \,\, \mathfrak{R}(b) \gt 0

## 3.267.1

\begin{align}
\int\limits_{0}^{1} x^{3n} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-2/3} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left( \frac{2}{3},n+\frac{1}{3} \right) \\
&= \frac{\Gamma\left(\frac{2}{3}\right) \Gamma\left(n + \frac{1}{3}\right)}{3\Gamma(n+1)}
= \frac{2\pi}{3\sqrt{3}} \frac{\Gamma\left(n + \frac{1}{3}\right)}{\Gamma\left(\frac{1}{3}\right) \Gamma(n+1)}
\end{align}

## 3.267.2

\begin{align}
\int\limits_{0}^{1} x^{3n-1} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-1} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left(n, \frac{2}{3} \right) \\
&= \frac{\Gamma(n)\Gamma(2/3)}{3\Gamma\left(n + \frac{2}{3}\right)}
= \frac{(n-1)! \Gamma(2/3)}{3\Gamma\left(n + \frac{2}{3}\right)}
\end{align}

## 3.267.3

\begin{align}
\int\limits_{0}^{1} x^{3n-2} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-4/3} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left( \frac{2}{3}, n \, – \frac{1}{3} \right) \\
&= \frac{\Gamma\left(n – \frac{1}{3}\right)\Gamma(2/3)}{3\Gamma\left(n + \frac{1}{3}\right)}
\end{align}

we used the substitution $$y=x^3$$ for all three integrals.