## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[3]{1+\sqrt[3]{x^{2}}}}{x \sqrt[9]{x^{8}}} d x
$$
## Solution

$$
\int \frac{\sqrt[3]{1+\sqrt[3]{x^{2}}}}{x \sqrt[9]{x^{8}}} d x=\int x^{-\frac{17}{9}}\left(1+x^{\frac{2}{3}}\right)^{\frac{1}{3}} d x=
$$

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

$$
m=-\frac{17}{9} ; n=\frac{2}{3} ; p=\frac{1}{3}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{17}{9}+1}{\frac{2}{3}}+\frac{1}{3}=-\frac{4}{3}+\frac{1}{3}=-1$ 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:

$$
\begin{aligned}
& x^{-\frac{2}{3}}+1=z^{3} \\
& x=\left(z^{3}-1\right)^{-\frac{3}{2}} \\
& d x=-\frac{3}{2}\left(z^{3}-1\right)^{-\frac{5}{2}} \cdot 3 z^{2} d z=-\frac{9}{2} z^{2}\left(z^{3}-1\right)^{-\frac{5}{2}} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{3}-1\right)^{-\frac{3}{2}}\right)^{-\frac{17}{9}} \cdot\left(1+\left(\left(z^{3}-1\right)^{-\frac{3}{2}}\right)^{\frac{2}{3}}\right)^{\frac{1}{3}}\left(-\frac{9}{2}\right) z^{2}\left(z^{3}-1\right)^{-\frac{5}{2}} d z= \\
& =-\frac{9}{2} \int\left(z^{3}-1\right)^{\frac{17}{6}} \cdot\left(1+\frac{1}{z^{3}-1}\right)^{\frac{1}{3}} z^{2}\left(z^{3}-1\right)^{-\frac{5}{2}} d z= \\
& =-\frac{9}{2} \int\left(z^{3}-1\right)^{\frac{1}{3}} \cdot\left(\frac{z^{3}}{z^{3}-1}\right)^{\frac{1}{3}} z^{2} d z=-\frac{9}{2} \int\left(z^{3}-1\right)^{\frac{1}{3}} \cdot \frac{z}{\left(z^{3}-1\right)^{\frac{1}{3}}} z^{2} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-\frac{9}{2} \int z^{3} d z=-\frac{9}{2} \cdot \frac{z^{4}}{4}+C=\frac{-9\left(\sqrt[3]{x^{-\frac{2}{3}}+1}\right)^{4}}{8}+C= \\
& =-\frac{9}{8}\left(\sqrt[3]{\frac{1}{\sqrt[3]{x^{2}}}+1}\right)^{4}+C=-\frac{9}{8}\left(\sqrt[3]{\frac{1+\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}}\right)^{4}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-5$ » Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last edited: 10:44, 9 July 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 13-6

## Material from PlusPi