## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[4]{\left(1+\sqrt[3]{x^{2}}\right)^{3}}}{x^{2} \cdot \sqrt[6]{x}} d x
$$
## Solution

$$
\int \frac{\sqrt[4]{\left(1+\sqrt[3]{x^{2}}\right)^{3}}}{x^{2} \cdot \sqrt[6]{x}} d x=\int x^{-\frac{13}{6}}\left(1+x^{\frac{2}{3}}\right)^{\frac{3}{4}} d x=
$$

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

$$
m=-\frac{13}{6} ; n=\frac{2}{3} ; p=\frac{3}{4}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{13}{6}+1}{\frac{2}{3}}+\frac{3}{4}=-\frac{7}{4}+\frac{3}{4}=-1$ is an integer, we use the substitution:

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

That is, in our case, the substitution has the form:

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

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{4}-1\right)^{-\frac{3}{2}}\right)^{-\frac{13}{6}} \cdot\left(1+\left(\left(z^{4}-1\right)^{-\frac{3}{2}}\right)^{\frac{2}{3}}\right)^{\frac{3}{4}}(-6) z^{3}\left(z^{4}-1\right)^{-\frac{5}{2}} d z= \\
& =-6 \int\left(z^{4}-1\right)^{\frac{13}{4}} \cdot\left(1+\frac{1}{z^{4}-1}\right)^{\frac{3}{4}} z^{3}\left(z^{4}-1\right)^{-\frac{10}{4}} d z= \\
& =-6 \int\left(z^{4}-1\right)^{\frac{3}{4}} \cdot\left(\frac{z^{4}}{z^{4}-1}\right)^{\frac{3}{4}} z^{3} d z=-6 \int\left(z^{4}-1\right)^{\frac{3}{4}} \cdot \frac{z^{3}}{\left(z^{4}-1\right)^{\frac{3}{4}}} z^{3} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-6 \int z^{6} d z=-6 \cdot \frac{z^{7}}{7}+C=\frac{-6\left(\sqrt[4]{x^{-\frac{2}{3}}+1}\right)^{7}}{7}+C= \\
& =-\frac{6}{7}\left(\sqrt[4]{\frac{1}{\sqrt[3]{x^{2}}}+1}\right)^{7}+C=-\frac{6}{7}\left(\sqrt[4]{\frac{1+\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}}\right)^{7}+C=-\frac{6 \cdot \sqrt[4]{\left(1+\sqrt[3]{x^{2}}\right)^{7}}}{7 \sqrt[6]{x^{7}}}+C
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_13-13" Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last modified: 15:59, 19 July 2009.
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## Problem Kuznetsov Integrals 13-14

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