## Condition of the problem

Find the indefinite integral:

$$
\int \frac{\sqrt[5]{\left(1+\sqrt[5]{x^{4}}\right)^{4}}}{x^{2} \cdot \sqrt[25]{x^{11}}} d x
$$
## Solution

$$
\int \frac{\sqrt[3]{\left(1+\sqrt[5]{x^{4}}\right)^{4}}}{x^{2} \cdot \sqrt[25]{x^{11}}} d x=\int x^{-\frac{61}{25}}\left(1+x^{\frac{4}{5}}\right)^{\frac{4}{5}} d x=
$$

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

$$
m=-\frac{61}{25} ; n=\frac{4}{5} ; p=\frac{4}{5}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{61}{25}+1}{\frac{4}{5}}+\frac{4}{5}=-\frac{9}{5}+\frac{4}{5}=-1$ is an integer, we use the substitution:

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

In our case, the substitution has the form:

$$
\begin{aligned}
& x^{-\frac{4}{8}}+1=z^{5} \\
& x=\left(z^{5}-1\right)^{-\frac{5}{4}} \\
& d x=-\frac{5}{4}\left(z^{5}-1\right)^{-\frac{2}{4}} \cdot 5 z^{4} d z=-\frac{25}{4} \cdot z^{4}\left(z^{5}-1\right)^{-\frac{9}{4}} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{5}-1\right)^{-\frac{5}{4}}\right)^{-\frac{61}{25}} \cdot\left(1+\left(\left(z^{5}-1\right)^{-\frac{5}{4}}\right)^{\frac{4}{5}}\right)^{\frac{4}{5}} \cdot\left(-\frac{25}{4}\right) z^{4}\left(z^{5}-1\right)^{-\frac{9}{4}} d z= \\
& =-\frac{25}{4} \cdot \int\left(z^{5}-1\right)^{\frac{61}{20}} \cdot\left(1+\frac{1}{z^{5}-1}\right)^{\frac{4}{5}} \cdot z^{4}\left(z^{5}-1\right)^{-\frac{9}{4}} d z= \\
& =-\frac{25}{4} \cdot \int\left(z^{5}-1\right)^{\frac{6}{20}} \cdot\left(\frac{z^{5}}{z^{5}-1}\right)^{\frac{5}{5}} \cdot z^{4}\left(z^{5}-1\right)^{-\frac{45}{20}} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-\frac{25}{4} \cdot \int\left(z^{5}-1\right)^{\frac{16}{20}} \cdot\left(\frac{z^{5}}{z^{5}-1}\right)^{\frac{4}{5}} \cdot z^{4} d z=-\frac{25}{4} \cdot \int\left(z^{5}-1\right)^{\frac{4}{5}} \cdot \frac{z^{4}}{\left(z^{5}-1\right)^{\frac{4}{5}}} \cdot z^{4} d z= \\
& =-\frac{25}{4} \cdot \int z^{8} d z=-\frac{25}{4} \cdot \frac{z^{9}}{9}+C=-\frac{25 z^{9}}{36}+C=-\frac{25\left(\sqrt[5]{x^{-\frac{4}{5}}+1}\right)^{9}}{36}+C= \\
& =-\frac{25}{36}\left(\sqrt[5]{\frac{1}{\sqrt[5]{x^{4}}}+1}\right)^{9}+C=-\frac{25}{36}\left(\sqrt[5]{\frac{1+\sqrt[5]{x^{4}}}{\sqrt[5]{x^{4}}}}\right)^{9}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-21 »$ Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last edited: 10:15, 22 February 2010.
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## Problem Kuznetsov Integrals 13-22

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