## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[4]{1+\sqrt[3]{x^{2}}}}{x \sqrt[6]{x^{5}}} d x
$$
## Solution

$$
\int \frac{\sqrt[4]{1+\sqrt[3]{x^{2}}}}{x \sqrt[6]{x^{5}}} d x=\int x^{-\frac{11}{6}}\left(1+x^{\frac{2}{3}}\right)^{\frac{1}{4}} d x=
$$

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

$$
m=-\frac{11}{6} ; n=\frac{2}{3} ; p=\frac{1}{4}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{11}{6}+1}{\frac{2}{3}}+\frac{1}{4}=-\frac{5}{4}+\frac{1}{4}=-1$ 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:

$$
\begin{aligned}
& x^{-\frac{2}{3}}+1=z^{4} \\
& x=\left(z^{4}-1\right)^{-\frac{3}{2}} \\
& d x=-\frac{3}{2}\left(z^{4}-1\right)^{-\frac{5}{2}} \cdot 4 z^{3} d z=-6 z^{3}\left(z^{4}-1\right)^{-\frac{5}{2}} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{4}-1\right)^{-\frac{3}{2}}\right)^{-\frac{11}{6}} \cdot\left(1+\left(\left(z^{4}-1\right)^{-\frac{3}{2}}\right)^{\frac{2}{3}}\right)^{\frac{1}{4}}(-6) z^{3}\left(z^{4}-1\right)^{-\frac{5}{2}} d z= \\
& =-6 \int\left(z^{4}-1\right)^{\frac{11}{4}} \cdot\left(1+\frac{1}{z^{4}-1}\right)^{\frac{1}{4}} z^{3}\left(z^{4}-1\right)^{-\frac{10}{4}} d z= \\
& =-6 \int\left(z^{4}-1\right)^{\frac{1}{4}} \cdot\left(\frac{z^{4}}{z^{4}-1}\right)^{\frac{1}{4}} z^{3} d z=-6 \int\left(z^{4}-1\right)^{\frac{1}{4}} \cdot \frac{z}{\left(z^{4}-1\right)^{\frac{1}{4}}} z^{3} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-6 \int z^{4} d z=-6 \cdot \frac{z^{5}}{5}+C=\frac{-6\left(\sqrt[4]{x^{-\frac{2}{3}}+1}\right)^{5}}{5}+C= \\
& =-\frac{6}{5}\left(\sqrt[4]{\frac{1}{\sqrt[3]{x^{2}}}+1}\right)^{5}+C=-\frac{6}{5}\left(\sqrt[4]{\frac{1+\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}}\right)^{5}+C=-\frac{6 \cdot \sqrt[4]{\left(1+\sqrt[3]{x^{2}}\right)^{5}}}{5 \sqrt[6]{x^{5}}}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-29$ " Categories: Kuznetsov's Problem Book Integrals Problem $13 \mid$ Integrals

- Last edited: 08:46, 23 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 13-30

## Material from PlusPi