Computing $\int \sqrt{1-\frac{1}{9}x^{-2/3}}dx$

Question

I was trying an exercise from a textbook that eventually ended up needing to evaluate an integral. I got stumped and kept going in circles. I tried Trig substitution, and I even tried a trick with integrating by parts to and making $dv=1$. Eventually I gave up and looked in the back of the book and saw the final answer was simpler than I was expecting...That's when I re-read the problem and realized I wrote a fraction wrong and the intended integral was actually indeed, quite straight forward.

But now I am curious about the incorrectly written integral I was stuck on and how to solve it. The integral is: $$\int \sqrt{1-\frac{1}{9}x^{-2/3}}dx$$

Wolfram Alpha says the answer is $\left( 1-\frac{1}{9x^{2/3}}\right)^{3/2} x +C$ but I'm not sure how to get that.

Answer 1

HINT

To begin with, let us rearrange things to make it easier to solve: \begin{align*} \int\sqrt{1 - \frac{1}{9x^{2/3}}}\mathrm{d}x & = \int\frac{1}{3x^{1/3}}\sqrt{9x^{{2/3}} - 1}\mathrm{d}x\\\\ & = \frac{1}{2}\int\frac{2}{3x^{1/3}}\sqrt{9x^{{2/3}} - 1}\mathrm{d}x\\\\ & = \frac{1}{2}\int\sqrt{9x^{2/3} - 1}(x^{2/3})'\mathrm{d}x\\\\ & = \frac{1}{2}\int\sqrt{9u - 1}\mathrm{d}u \end{align*}

Can you take it from here?

Answer 2

The change of variable $t=x^{-2/3}$, $dt=-\tfrac{2}{3}x^{-5/3}\,dx$, yields $$I=\int \sqrt{1-\frac{1}{9}x^{-3/2}}dx=-\tfrac32\int\Big(1-\tfrac19 t\Big)^{1/2}t^{-5/2}\,dt=-\tfrac32\int\Big(\frac{1-\tfrac19 t}{t}\Big)^{1/2}t^{-2}\,dt $$

The change of variables $z=\Big(\frac{1-\tfrac19 t}{t}\Big)^{1/2}$, $t=\frac{1}{z^2+\tfrac19}$, $dt=-\frac{2z}{\big(z^2+\tfrac19\big)^2}dz$, yields \begin{align} I&=\tfrac13\int z^2\,dz=z^3+C\\ &=\Big(\frac{1-\tfrac19 x^{-2/3}}{x^{-2/3}}\Big)^{3/2} +C\\ &=(1-\tfrac19 x^{-2/3})^{3/2}x +C \end{align}

---

Comment: The integral of the OP is a type of integrals called differential binomials and were studied by P. L. Chebyshev:

$$ \int x^m(a+bx^n)^p\,dx\qquad a,b\in\mathbb{R}, \,m, n, p\in\mathbb{Q}$$ The change of variables $x=t^{1/n}$ yields $$ \int x^m(a+bx^n)^p\,dx = \frac1n\int (a+bt)^p t^{\tfrac{m+1}{n}-1}\,dt $$ There are three cases that are solvable

1. $p\in\mathbb{Z}$: If $q=\frac{m+1}{n}-1=\frac{\alpha}{\beta}$, $\alpha,\beta\in\mathbb{Z}$ and $g.c.d(\alpha,\beta)=1$, then the substitution $z=t^{1/\beta}$ reduced the integral to one of a rational fraction.

2. $q\in\mathbb{Z}$: the substitution $z=(a+bt)^{1/\beta}$ reduced to integral to one of a rational fraction.

3. $p+q\in\mathbb{Z}$. The change of variable $z=\Big(\frac{a+bt}{t}\Big)^{1/\beta}$ reduces the integral to one of a rational fraction.

If none of the cases 1-3 hold, then the integral cannot be expressed in terms of elementary functions.

More detail are in Kudriatzev, L. D., Curso de Análisis Mathematico, Vol. I (In Spanish) pp. 443-444, Mir, Moscow, 1983( Translation from Russian, MIR 1981 USSR).

The integral of the OP satisfies condition 3.