How to solve $\int\sqrt{1+x^{-2/3}}$ without integration by parts formula?

Question

Original question is to find the arc length of graph of $$y=\frac{3}{2}x^{2/3}+1,\frac{1}{8}\leq x\leq 1$$ Here is my solution: $$y'=x^{-1/3}$$ $$y'^2=x^{-2/3}$$ $$1+y'^2=1+x^{-2/3}$$ then arc length is $$L=\int_{1/8}^1\sqrt{1+f'^2}dx=\int_{1/8}^1\sqrt{1+x^{-2/3}}dx$$ but I cannot find its integral. I tries to use substitution with $u=x^{-2/3}$ and $u=x^{-1/3}$, but they didn't work. sadly...

so how to construct function $u$ to substitute integrand? or I have to use other method?

Any help will appreciated.

Answer 1

Let's solve $$I = \int\sqrt{1+x^{-2/3}}\, \mathrm{d}x$$ without using integration by parts. I shall leave the details for you to verify. First, substitute $u = x^{1/3}$ to get $$I = \int\sqrt{1+x^{-2/3}}\, \mathrm{d}x = 3\int u \sqrt{u^2 + 1}\, \mathrm{d}u$$ Now, put $v = u^2 + 1$ to get $$I = \frac{3}{2}\int \sqrt{v}\, \mathrm{d}v$$ which is easily seen to be $$I = v^{3/2} + C$$ Plugging in all the substitutions made above, we have $$I = (1 + x^{2/3})^{3/2} + C$$

Answer 2

It is very much possible through u-sub:

$$\large{\int_{1/8}^{1}{\sqrt{1+\dfrac{1}{x^{2/3}}}} \space \mathrm{dx} = \int_{1/8}^{1}{\dfrac{\sqrt{1 + x^{2/3}}}{|x^{1/3}|}}\space \mathrm{dx} \\ \quad \\ \quad \\ \xrightarrow{\Large{x \space > \space 0}} \quad \int_{1/8}^{1}{\sqrt{1 + x^{2/3}}\ \cdot \dfrac{1}{x^{1/3}}} \space \mathrm{dx} \\ \quad \\ \quad \\ \xrightarrow{ \Large{\begin{cases} u \space = \space 1 + x^{2/3} \\ du \space = \space \frac{2}{3x^{\Large{1/3}}} dx \end{cases}}} \quad \frac{3}{2}\int_{5/4}^{2}{\sqrt{u}} \space \mathrm{du}}$$

I believe the rest is clear.