## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[3]{1+\sqrt[5]{x^{4}}}}{x^{2} \cdot \sqrt[15]{x}} d x
$$
## Solution

$$
\int \frac{\sqrt[3]{1+\sqrt[5]{x^{4}}}}{x^{2} \cdot \sqrt[15]{x}} d x=\int x^{-\frac{31}{15}}\left(1+x^{\frac{4}{5}}\right)^{\frac{1}{3}} d x=
$$

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

$$
m=-\frac{31}{15} ; n=\frac{4}{5} ; p=\frac{1}{3}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{31}{15}+1}{\frac{4}{5}}+\frac{1}{3}=-\frac{4}{3}+\frac{1}{3}=-1$ is an integer, we use the substitution:

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

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

$$
\begin{aligned}
& x^{-\frac{4}{5}}+1=z^{3} \\
& x=\left(z^{3}-1\right)^{-\frac{5}{4}} \\
& d x=-\frac{5}{4}\left(z^{3}-1\right)^{-\frac{9}{4}} \cdot 3 z^{2} d z=-\frac{15}{4} \cdot z^{2}\left(z^{3}-1\right)^{-\frac{9}{4}} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{3}-1\right)^{-\frac{5}{4}}\right)^{-\frac{31}{15}} \cdot\left(1+\left(\left(z^{3}-1\right)^{-\frac{5}{4}}\right)^{\frac{4}{5}}\right)^{\frac{1}{3}} \cdot\left(-\frac{15}{4}\right) z^{2}\left(z^{3}-1\right)^{-\frac{9}{4}} d z= \\
& =-\frac{15}{4} \cdot \int\left(z^{3}-1\right)^{\frac{31}{12}} \cdot\left(1+\frac{1}{z^{3}-1}\right)^{\frac{1}{3}} \cdot z^{2}\left(z^{3}-1\right)^{-\frac{9}{4}} d z= \\
& =-\frac{15}{4} \cdot \int\left(z^{3}-1\right)^{\frac{1}{3}} \cdot\left(\frac{z^{3}}{z^{3}-1}\right)^{\frac{1}{3}} \cdot z^{2} \cdot d z=-\frac{15}{4} \cdot \int z^{3} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-\frac{15}{4} \cdot \frac{z^{4}}{4}+C=-\frac{15 z^{4}}{16}+C=-\frac{15\left(\sqrt[3]{x^{-\frac{4}{5}}+1}\right)^{4}}{16}+C= \\
& =-\frac{15}{16} \cdot\left(\sqrt[3]{\frac{1}{\sqrt[5]{x^{4}}}+1}\right)^{4}+C=-\frac{15}{16} \cdot\left(\sqrt[3]{\frac{1+\sqrt[5]{x^{4}}}{\sqrt[5]{x^{4}}}}\right)^{4}+C=-\frac{15\left(1+\sqrt[3]{x^{4}}\right)^{\frac{4}{3}}}{16 \sqrt[18]{x^{16}}}+C
\end{aligned}
$$

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_13-23" Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last edited: 09:48, 25 March 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 13-24

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