## Problem Statement

Find the indefinite integral:

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

$$
\int \frac{\sqrt[5]{(1+\sqrt[3]{x})^{4}}}{x \sqrt[5]{x^{3}}} d x=\int x^{-\frac{8}{5}}\left(1+x^{\frac{1}{3}}\right)^{\frac{4}{3}} d x=
$$

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

$$
m=-\frac{8}{5} ; n=\frac{1}{3} ; p=\frac{4}{5}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{8}{5}+1}{\frac{1}{3}}+\frac{4}{5}=-\frac{9}{5}+\frac{4}{5}=-1$ 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:

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

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{5}-1\right)^{-3}\right)^{-\frac{8}{5}} \cdot\left(1+\left(\left(z^{5}-1\right)^{-3}\right)^{\frac{1}{3}}\right)^{\frac{4}{3}} \cdot(-15) z^{4}\left(z^{5}-1\right)^{-4} d z= \\
& =-15 \cdot \int\left(z^{5}-1\right)^{\frac{24}{5}} \cdot\left(1+\frac{1}{z^{5}-1}\right)^{\frac{4}{5}} \cdot z^{4}\left(z^{5}-1\right)^{-4} d z= \\
& =-15 \cdot \int\left(z^{5}-1\right)^{\frac{24}{5}} \cdot\left(\frac{z^{5}}{z^{5}-1}\right)^{\frac{4}{5}} \cdot z^{4}\left(z^{5}-1\right)^{-4} d z= \\
& =-15 \cdot \int\left(z^{5}-1\right)^{4} \cdot z^{4}\left(z^{5}-1\right)^{-4} d z=-15 \cdot \int z^{8} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-15 \cdot \frac{z^{9}}{9}+C=-\frac{5 z^{9}}{3}+C=-\frac{5\left(\sqrt[5]{x^{-\frac{1}{3}}+1}\right)^{9}}{3}+C= \\
& =-\frac{5}{3}\left(\sqrt[5]{\frac{1}{\sqrt[3]{x}}+1}\right)^{9}+C=-\frac{5}{3}\left(\sqrt[5]{\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}}}\right)^{9}+C
\end{aligned}
$$

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_13-18" Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last modified: 11:58, 8 November 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 13-19

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