Example 3.10. Find the integral $\int \frac{d x}{\sqrt{2 x+3}+\sqrt[3]{(2 x+3)^{2}}}$.
Solution. Note that $\sqrt{2 x+3}=(2 x+3)^{\frac{1}{2}}, \sqrt[3]{(2 x+3)^{2}}=(2 x+3)^{\frac{2}{3}}$. The least common multiple of the denominators of the fractions $\frac{1}{2}, \frac{2}{3}$ is 6. Therefore, if we apply the substitution $2 x+3=t^{6}$, we will have:

$$
t=(2 x+3)^{\frac{1}{6}}, \sqrt{2 x+3}=\left(t^{6}\right)^{\frac{1}{2}}=t^{3}, \sqrt[3]{(2 x+3)^{2}}=\left(t^{6}\right)^{\frac{2}{3}}=t^{4}
$$

i.e., the irrationalities in the integrand expression disappear. Since: $x=\frac{1}{2}\left(t^{6}-3\right)$, then $d x=\frac{1}{2} 6 t^{5} d t=3 t^{5} d t$.

Substituting the found expressions into the desired integral, we get:

$$
\int \frac{d x}{\sqrt{2 x+3}+\sqrt[3]{(2 x+3)^{2}}}=\int \frac{3 t^{5} d t}{t^{3}+t^{4}}=3 \int \frac{t^{2} d t}{1+t} .
$$

Thus, the given integral is reduced to the integral of a rational function. To find it, we separate the integer part of the integrand function:

$$
\frac{t^{2}}{1+t}=\frac{t^{2}-1+1}{1+t}=\frac{(t-1)(t+1)+1}{t+1}=t-1+\frac{1}{t+1}
$$

Integrating each of the terms, we find:

$$
\int \frac{t^{2} d t}{1+t}=\frac{t^{2}}{2}-t+\ln |t+1|+C
$$

Returning to the original variable. Since $t=\sqrt[6]{2 x+3}$, we obtain the following final result:

$$
\int \frac{d x}{\sqrt{2 x+3}+\sqrt[3]{(2 x+3)^{2}}}=\frac{3}{2} \sqrt[3]{2 x+3}-3 \sqrt[6]{2 x+3}+3 \ln |\sqrt[6]{2 x+3}+1|+C
$$