1184. Find the integrals:
1) $\int \frac{d x}{\sqrt{x}+\sqrt[4]{x}}$
2) $\int \frac{x+\sqrt{1+x}}{\sqrt[3]{1+x}} d x$
3) $\int \frac{\sqrt{x}}{x^{2} \sqrt{x-1}} d x$
Solution. 1) Apply the substitution (a):

$$
\sqrt[4]{x}=t, \text { from which } x=t^{4}, d x=4 t^{3} d t
$$

$$
\int \frac{d x}{\sqrt{x}+\sqrt[4]{x}}=\int \frac{4 t^{3} d t}{t^{2}+t}=4 \int \frac{t^{2} d t}{t+1}=4 \int\left(t-1+\frac{1}{t+1}\right) d t=4 \times
$$

$$
\times\left[\frac{t^{2}}{2}-t+\ln (t+1)\right]+C=4\left[\frac{\sqrt{x}}{2}-\sqrt[4]{x}+\ln (1+\sqrt[4]{x})\right]+C.
$$

2) Apply the substitution (b). The least common multiple of the exponents of the roots in the integrand $m=6$, hence,

$$
t=\sqrt[6]{1+x}, \text { from which } 1+x=t^{6}, x=t^{6}-1, d x=6 t^{5} d t \text { and }
$$

$$
\int \frac{x+\sqrt{1+x}}{\sqrt[3]{1+x}} d x=\int \frac{t^{6}-1+t^{3}}{t^{2}} 6 t^{5} d t=6 \int\left(t^{9}+t^{6}-t^{9}\right) d t=
$$

$$
\begin{gathered}
=6\left(\frac{t^{10}}{10}+\frac{t^{7}}{7}-\frac{t^{4}}{4}\right)+C=6 \sqrt[3]{(1+x)^{2}}\left(\frac{1+x}{10}+\frac{\sqrt{1+x}}{7}-\right. \\
\left.-\frac{1}{4}\right)+C .
\end{gathered}
$$

3) Apply the substitution (a):

$$
t=\sqrt{\frac{x}{x-1}}, \text { from which } x=\frac{t^{2}}{t^{2}-1}, \quad d x=\frac{-2 t}{\left(t^{2}-1\right)^{2}} d t
$$

$$
\int \frac{\sqrt{x}}{x^{2} \sqrt{x-1}} d x=\int \frac{1}{x^{2}} \sqrt{\frac{x}{x-1}} d x=-\int \frac{t\left(t^{2}-1\right)^{2} 2 t}{t^{4}\left(t^{2}-1\right)^{2}} d t=
$$

$$
\begin{gathered}
=-\int \frac{2}{t^{2}} d t=-2 \int t^{-2} d t=-2 \frac{t^{-1}}{-1}+C= \\
=\frac{2}{t}+C=2 \sqrt{\frac{x-1}{x}}+C
\end{gathered}
$$

Verification. $F(x)=2 \sqrt{\frac{x-1}{x}}, F^{\prime}(x)=2 \frac{1}{2 \sqrt{\frac{x-1}{x}}} \times$

$$
\times \frac{x-(x-1)}{x^{2}}=\frac{1}{x^{2}} \sqrt{\frac{x}{x-1}}, \quad \text { therefore, the integration is correct. }
$$

Find the integrals: