## Problem Statement

Find the indefinite integral:

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

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

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

$$
m=-\frac{11}{5} ; n=\frac{4}{5} ; p=\frac{1}{2}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{11}{5}+1}{\frac{4}{5}}+\frac{1}{2}=-\frac{3}{2}+\frac{1}{2}=-1$ 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:

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

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{2}-1\right)^{-\frac{5}{4}}\right)^{-\frac{11}{5}} \cdot\left(1+\left(\left(z^{2}-1\right)^{-\frac{5}{4}}\right)^{\frac{4}{5}}\right)^{\frac{1}{2}} \cdot\left(-\frac{5}{2}\right) z\left(z^{2}-1\right)^{-\frac{9}{4}} d z= \\
& =-\frac{5}{2} \cdot \int\left(z^{2}-1\right)^{\frac{11}{4}} \cdot\left(1+\frac{1}{z^{2}-1}\right)^{\frac{1}{2}} \cdot z\left(z^{2}-1\right)^{-\frac{9}{4}} d z= \\
& =-\frac{5}{2} \cdot \int\left(z^{2}-1\right)^{\frac{1}{2}} \cdot\left(\frac{z^{2}}{z^{2}-1}\right)^{\frac{1}{2}} \cdot z \cdot d z=-\frac{5}{2} \cdot \int z^{2} d z=
\end{aligned}
$$

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

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}+13-22 »

Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

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## Problem Kuznetsov Integrals 13-23

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