Example 2. Find the integral $\int \sqrt[3]{\frac{2-x}{2+x}} \cdot \frac{1}{(2-x)^{2}} d x$.
Solution. This is an integral of the form (4.32), for which $\frac{a x+b}{c x+d}=\frac{2-x}{2+x}$, $\frac{p_{1}}{q_{1}}=\frac{1}{3}$

We apply the substitution

$$
\frac{2-x}{2+x}=t^{3}
$$

Express $x, 2-x$ and $d x$ in terms of the new variable $t$:

$$
\begin{gathered}
2-x=t^{3}(2+x), 2-2 t^{3}=x+x t^{3}, 2-2 t^{3}=x\left(1+t^{3}\right) \\
x=\frac{2-2 t^{3}}{1+t^{3}}, 2-x=2-\frac{2-2 t^{3}}{1+t^{3}}=\frac{4 t^{3}}{1+t^{3}}, \frac{1}{(2-x)^{2}}=\frac{\left(1+t^{3}\right)^{2}}{16 t^{6}} \\
d x=\frac{-6 t^{2}\left(1+t^{3}\right)-3 t^{2}\left(2-2 t^{3}\right)}{\left(1+t^{3}\right)^{2}} d t=\frac{-12 t^{2}}{\left(1+t^{3}\right)^{2}} d t
\end{gathered}
$$

Substituting the found expressions into the integral, we get

$$
\begin{gathered}
\int \sqrt[3]{\frac{2-x}{2+x}} \cdot \frac{1}{(2-x)^{2}} d x=\int t \cdot \frac{\left(1+t^{3}\right)^{2}}{16 t^{6}} \cdot \frac{\left(-12 t^{2}\right)}{\left(1+t^{3}\right)^{2}} d t=-\frac{3}{4} \int \frac{d t}{t^{3}}= \\
=\frac{3}{8 t^{2}}+C=\frac{3}{8} \sqrt[3]{\left(\frac{2+x}{2-x}\right)^{2}}+C
\end{gathered}
$$