Example 7. Find $\int \frac{\sqrt{1+\sqrt[3]{x}}}{x \sqrt{x}} d x$.
Solution. Let's represent the integral in the standard form $\int x^{-3 / 2}\left(1+x^{1 / 3}\right)^{1 / 2} d x . \quad$ Since $\quad m=-\frac{3}{2}, \quad n=\frac{1}{3}, \quad p=\frac{1}{2} \quad$ and $\frac{m+1}{n}+p=-1-$ is an integer, we will perform the third substitution of Chebyshev $(a=1, b=1): x^{-1 / 3}+1=t^{2}, \quad t=\sqrt{1+x^{-1 / 3}}$ From this, $\quad x=\frac{1}{\left(t^{2}-1\right)^{3}}, \quad d x=-\frac{6 t d t}{\left(t^{2}-1\right)^{4}}, \quad x^{-3 / 2}=\left(t^{2}-1\right)^{9 / 2}$, $1+x^{1 / 3}=1+\frac{1}{t^{2}-1}=\frac{t^{2}}{t^{2}-1}$. Then, after the substitutions in the desired integral, we get sequentially:

$$
\begin{aligned}
& \int \frac{\sqrt{1+\sqrt[3]{x}}}{x \sqrt{x}} d x=-6 \int \frac{t}{\sqrt{t^{2}-1}} \cdot\left(t^{2}-1\right)^{9 / 2} \cdot \frac{t}{\left(t^{2}-1\right)^{4}} d t= \\
& =-6 \int t^{2} d t=-2 t^{3}+C=-2 \sqrt{\left(1+x^{-1 / 3}\right)^{3}}+C
\end{aligned}
$$