Example 3. Find the integral $\int \frac{\sqrt{x+1}+2}{(x+1)^{2}-\sqrt{x+1}} d x$.
Solution. Here $\frac{a x+b}{c x+d}=x+1, \frac{p_{1}}{q_{1}}=\frac{1}{2}$. Let

$$
x+1=t^{2}
$$

then

$$
\begin{aligned}
d x & =2 t d t \\
\int \frac{\sqrt{x+1}+2}{(x+1)^{2}-\sqrt{x+1}} d x & =2 \int \frac{t+2}{t^{4}-t} t d t=2 \int \frac{t+2}{t^{3}-1} d t
\end{aligned}
$$

Decomposing the integrand into partial fractions, we get

$$
\frac{t+2}{t^{3}-1}=\frac{1}{t-1}-\frac{t+1}{t^{2}+t+1}
$$

Therefore,

$$
\begin{gathered}
2 \int \frac{t+2}{t^{3}-1} d t=2 \int\left(\frac{1}{t-1}-\frac{t+1}{t^{2}+t+1}\right) d t=2 \int \frac{d t}{t-1}- \\
-2 \int \frac{t+1}{t^{2}+2 \cdot \frac{1}{2} t+\frac{1}{4}-\frac{1}{4}+1} d t=2 \int \frac{d(t-1)}{t-1}- \\
-2 \int \frac{\left(t+\frac{1}{2}\right)-\frac{1}{2}+1}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}} d\left(t+\frac{1}{2}\right)=2 \int \frac{d(t-1)}{t-1}- \\
-2 \int \frac{\left(t+\frac{1}{2}\right) d\left(t+\frac{1}{2}\right)}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}-\int \frac{d\left(t+\frac{1}{2}\right)}{\left(t+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=2 \ln |t-1|- \\
-\ln \left[\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}\right]-\frac{2}{\sqrt{3}} \operatorname{arctg} \frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}+C=\ln (t-1)^{2}- \\
-
\end{gathered}
$$

Solution. This is also an integral of type (4.32), for which $\frac{a x+b}{c x+d}=x$ (i.e., $a=1, b=0, c=0, d=1$), $\frac{p_{1}}{q_{1}}=\frac{1}{3}, \frac{p_{2}}{q_{2}}=\frac{2}{3}, \frac{p_{3}}{q_{3}}=\frac{1}{6}$, the common denominator of all fractions is 6, so we apply the substitution
$x=t^{6}$ (which allows us to get rid of all radicals).

Since

$$
x=t^{6}
$$

then

$$
d x=6 t^{5} d t, \sqrt[3]{x}=t^{2}, \sqrt[3]{x^{2}}=t^{4}
$$

Therefore,

$$
\begin{aligned}
& \qquad \begin{array}{l}
\int \frac{x+\sqrt[3]{x^{2}}+\sqrt[6]{x}}{x(1+\sqrt[3]{x})} d x=\int \frac{t^{6}+t^{4}+t}{t^{6}\left(1+t^{2}\right)} 6 t^{5} d t=6 \int \frac{t^{5}+t^{3}+1}{1+t^{2}} d t= \\
=6 \int \frac{t^{3}\left(t^{2}+1\right)+1}{1+t^{2}} d t=6 \int t^{3} d t+6 \int \frac{d t}{1+t^{2}}=\frac{3}{2} t^{4}+6 \operatorname{arctg} t+ \\
+C=\frac{3}{2} \sqrt[3]{x^{2}}+6 \operatorname{arctg} \sqrt[6]{x}+C .
\end{array} \\
& \text { Example } 5 . \text { Find the integral } \int \frac{\sqrt{1+\sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x .
\end{aligned}
$$

Solution. Rewriting the integral as

$$
\int x^{-\frac{2}{3}}\left(1+x^{\frac{1}{3}}\right)^{\frac{1}{2}} d x
$$

and comparing it with the integral (4.33), we conclude that $m=-\frac{2}{3}, n=\frac{1}{3}, p=\frac{1}{2}$.

Since

$$
\frac{m+1}{n}=\frac{-\frac{2}{3}+1}{\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{1}{3}}=1
$$

is an integer, we have the second case of integrability of the differential binomial.

The substitution $a+b x^{n}=t^{s}$ in this case will be

$$
\left(1+x^{\frac{1}{3}}\right)=t^{2}
$$

from which

$$
x^{\frac{1}{3}}=t^{2}-1, \frac{1}{3} x^{-\frac{2}{3}} d x=2 t d t, x^{-\frac{2}{3}} d x=6 t d t
$$

Substituting these expressions into the integral, we get

$$
\begin{gathered}
\int x^{-\frac{2}{3}}\left(1+x^{\frac{1}{3}}\right)^{\frac{1}{2}} d x=\int\left(1+x^{\frac{1}{3}}\right)^{\frac{1}{2}} x^{-\frac{2}{3}} d x=\int t \cdot 6 t d t= \\
=6 \int t^{2} d t=6 \cdot \frac{t^{3}}{3}+C=2\left(1+x^{\frac{1}{3}}\right)^{\frac{3}{2}}+C
\end{gathered}
$$