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For this problem I did decided to use a simple $u$ substitution of $u = \ln x$ .

So, using $u = \ln x$ and the change of interval from $0$ and $1$ we have:

$\begin{aligned} &\int^1_0 \frac{du}{u^{\frac{1}{3}}} \\ &\implies \left[\frac{3}{2}u^{\frac{2}{3}}\right]^1_0 \\ &\implies \boxed{\frac{3}{2}} \end{aligned}$

One should observe that this is an improper integral, undefined at the lower limit of integration, since $\ln(1)=0$ . To do the problem properly, use Jeel's method of substitution, $u=\ln(x)$ , to find the integral from $1+h$ to $e$ , and then let $h$ approach zero from above.

let ln(x)=t^3 so (1/x)dx=3(t^2)dt put it in eq. and solve it.

$\begin{aligned} \int_{1}^{e} \frac{dx}{x\sqrt[3]{\ln(x)}} &= \frac{3}{2} \ln^{\frac{2}{3}}(x) \color{grey}{+C} \quad\quad\quad\quad\quad{\text{Indefinite Integral}} \\&= \frac{3}{2} \space \square \end{aligned}$

$\large \text{ADIOS!!!}$

Let $u=\ln x\Leftrightarrow du=\frac{1}{x}dx$ and $u\left( e \right)=\ln \left( e \right)=1\,\,\And \,\,u\left( 1 \right)=\ln \left( 1 \right)=0$

So $\int_{1}^{e}{\frac{dx}{x{{\left( \ln x \right)}^{1/3}}}=\int_{0}^{1}{\frac{du}{{{u}^{1/3}}}}=\int_{0}^{1}{{{u}^{-1/3}}du}=\frac{3}{2}\left[ {{u}^{2/3}} \right]_{0}^{1}=\frac{3}{2}=1.5}$

put Ln(x)=t^3.....change the limits and the solution can be visible easily