## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[5]{\left(1+\sqrt[4]{x^{3}}\right)^{4}}}{x^{2} \cdot \sqrt[20]{x^{7}}} d x
$$
## Solution

$$
\int \frac{\sqrt[5]{\left(1+\sqrt[4]{x^{3}}\right)^{4}}}{x^{2} \cdot \sqrt[20]{x^{7}}} d x=\int x^{-\frac{47}{20}}\left(1+x^{\frac{3}{4}}\right)^{\frac{4}{5}} d x=
$$

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

$$
m=-\frac{47}{20} ; n=\frac{3}{4} ; p=\frac{4}{5}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{47}{20}+1}{\frac{3}{4}}+\frac{4}{5}=-\frac{9}{5}+\frac{4}{5}=-1$ 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:

$$
\begin{aligned}
& x^{-\frac{3}{4}}+1=z^{5} \\
& x=\left(z^{5}-1\right)^{-\frac{4}{3}} \\
& d x=-\frac{4}{3}\left(z^{5}-1\right)^{-\frac{5}{3}} \cdot 5 z^{4} d z=-\frac{20}{3} \cdot z^{4}\left(z^{5}-1\right)^{-\frac{5}{3}} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{5}-1\right)^{-\frac{4}{3}}\right)^{-\frac{47}{20}} \cdot\left(1+\left(\left(z^{5}-1\right)^{-\frac{4}{3}}\right)^{\frac{3}{4}}\right)^{\frac{4}{5}} \cdot\left(-\frac{20}{3}\right) z^{4}\left(z^{5}-1\right)^{-\frac{7}{3}} d z= \\
& =-\frac{20}{3} \cdot \int\left(z^{5}-1\right)^{\frac{47}{15}} \cdot\left(1+\frac{1}{z^{5}-1}\right)^{\frac{4}{5}} \cdot z^{4}\left(z^{5}-1\right)^{-\frac{7}{3}} d z= \\
& =-\frac{20}{3} \cdot \int\left(z^{5}-1\right)^{\frac{47}{15}} \cdot\left(\frac{z^{5}}{z^{5}-1}\right)^{\frac{4}{5}} \cdot z^{4}\left(z^{5}-1\right)^{-\frac{35}{15}} d z=-\frac{20}{3} \cdot \int z^{8} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-\frac{20}{3} \cdot \frac{z^{9}}{9}+C=-\frac{20 z^{9}}{27}+C=-\frac{20\left(\sqrt[5]{x^{-\frac{3}{4}}+1}\right)^{9}}{27}+C= \\
& =-\frac{20}{27}\left(\sqrt[5]{\frac{1}{\sqrt[4]{x^{3}}}+1}\right)^{9}+C=-\frac{20}{27}\left(\sqrt[5]{\frac{1+\sqrt[4]{x^{3}}}{\sqrt[4]{x^{3}}}}\right)^{9}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-20$ » Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last edited: 12:34, 8 November 2009.
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## Problem Kuznetsov Integrals 13-21

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