Example 1.18. Find the integral $\int x^{1 / 3}\left(2+3 x^{1 / 2}\right)^{3} d x$.
Solution. Here $m=1 / 3, n=1 / 2, p=3$. The least common multiple of the denominators of $m$ and $n$, i.e., the numbers 2 and 3, is 6. Therefore, we set $t=\sqrt[6]{x}$. Then $x=t^{6}, d x=6 t^{5} d t$, $x^{1 / 3}=t^{2}, x^{1 / 2}=t^{3}$. As a result of applying the specified substitution, the integral is transformed as follows:

$$
\begin{gathered}
\int x^{1 / 3}\left(2+3 x^{1 / 2}\right)^{3} d x=6 \int t^{2}\left(2+3 t^{2}\right)^{3} t^{5} d t= \\
=6 \int t^{7}\left(8+36 t^{3}+54 t^{6}+27 t^{9}\right) d t=
\end{gathered}
$$

$$
\begin{gathered}
=48 \int t^{7} d t+216 \int t^{10} d t+234 \int t^{13} d t+162 \int t^{16} d t= \\
\quad=\frac{48}{8} t^{8}+\frac{216}{11} t^{11}+\frac{324}{14} t^{14}+\frac{162}{17} t^{17}+C= \\
=6 \sqrt[3]{x^{4}}+\frac{216}{11} \sqrt[6]{x^{11}}+\frac{162}{7} \sqrt[3]{x^{7}}+\frac{162}{17} \sqrt[6]{x^{17}}+C
\end{gathered}
$$

2) The number $\frac{m+1}{n}$ is an integer. Then the integral $\int x^{m}\left(a+b x^{n}\right)^{p} d x$ is rationalized by the substitution $t=\sqrt[3]{a x+b x^{n}}$, where $s-$ is the denominator of the rational number $p$.