## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt{1+\sqrt[3]{x}}}{x \sqrt{x}} d x
$$
## Solution

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

Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which

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

Since $\frac{m+1}{n}+p=\frac{-\frac{3}{2}+1}{\frac{1}{3}}+\frac{1}{2}=-\frac{3}{2}+\frac{1}{2}=-1$ is an integer, we use the substitution:

$a x^{-n}+b=z^{s}$, where $s$ is the denominator of the fraction $p$.

In our case, the substitution is:

$$
\begin{aligned}
& x^{-\frac{1}{3}}+1=z^{2} \\
& x=\left(z^{2}-1\right)^{-3} \\
& d x=-3\left(z^{2}-1\right)^{-4} \cdot 2 z d z=-6 z\left(z^{2}-1\right)^{-4} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{2}-1\right)^{-3}\right)^{-\frac{3}{2}} \cdot\left(1+\left(\left(z^{2}-1\right)^{-3}\right)^{\frac{1}{3}}\right)^{\frac{1}{2}}(-6) z\left(z^{2}-1\right)^{-4} d z= \\
& =-6 \int\left(z^{2}-1\right)^{\frac{9}{2}} \cdot\left(1+\frac{1}{z^{2}-1}\right)^{\frac{1}{2}} z\left(z^{2}-1\right)^{-4} d z= \\
& =-6 \int\left(z^{2}-1\right)^{\frac{1}{2}} \cdot\left(\frac{z^{2}}{z^{2}-1}\right)^{\frac{1}{2}} z d z=-6 \int\left(z^{2}-1\right)^{\frac{1}{2}} \cdot \frac{z}{\left(z^{2}-1\right)^{\frac{1}{2}}} z d z= \\
& =-6 \int z^{2} d z=-6 \cdot \frac{z^{3}}{3}+C=-2 z^{3}+C=-2\left(\sqrt{x^{-\frac{1}{3}}+1}\right)^{3}+C=
\end{aligned}
$$

$$
=-2 \sqrt{\left(\frac{1}{\sqrt[3]{x}}+1\right)^{3}}+C=-2 \sqrt{\frac{(1+\sqrt[3]{x})^{3}}{x}}+C
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-3$ »

Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

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