## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[5]{(1+\sqrt{x})^{4}}}{x \sqrt[10]{x^{9}}} d x
$$
## Solution

$$
\int \frac{\sqrt[8]{(1+\sqrt{x})^{4}}}{x \sqrt[10]{x^{9}}} d x=\int x^{-\frac{19}{10}}\left(1+x^{\frac{1}{2}}\right)^{\frac{4}{8}} d x=
$$

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

$$
m=-\frac{19}{10} ; n=\frac{1}{2} ; p=\frac{4}{5}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{19}{10}+1}{\frac{1}{2}}+\frac{4}{5}=-\frac{9}{5}+\frac{4}{5}=-1$ 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:

$$
\begin{aligned}
& x^{-\frac{1}{2}}+1=z^{5} \\
& x=\left(z^{5}-1\right)^{-2} \\
& d x=-2\left(z^{5}-1\right)^{-3} \cdot 5 z^{4} d z=-10 z^{4}\left(z^{5}-1\right)^{-3} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{5}-1\right)^{-2}\right)^{-\frac{19}{10}} \cdot\left(1+\left(\left(z^{5}-1\right)^{-2}\right)^{\frac{1}{2}}\right)^{\frac{4}{5}} \cdot(-10) z^{4}\left(z^{5}-1\right)^{-3} d z= \\
& =-10 \cdot \int\left(z^{5}-1\right)^{\frac{19}{5}} \cdot\left(1+\frac{1}{z^{5}-1}\right)^{\frac{4}{5}} \cdot z^{4}\left(z^{5}-1\right)^{-3} d z= \\
& =-10 \cdot \int\left(z^{5}-1\right)^{\frac{19}{5}} \cdot\left(\frac{z^{5}}{z^{5}-1}\right)^{\frac{4}{5}} \cdot z^{4}\left(z^{5}-1\right)^{-3} d z= \\
& =-10 \cdot \int\left(z^{5}-1\right)^{\frac{15}{5}} \cdot z^{4} \cdot z^{4}\left(z^{5}-1\right)^{-3} d z=-10 \cdot \int z^{8} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-10 \cdot \frac{z^{9}}{9}+C=-\frac{10 z^{9}}{9}+C=-\frac{10\left(\sqrt[5]{x^{-\frac{1}{2}}+1}\right)^{9}}{9}+C= \\
& =-\frac{10}{9}\left(\sqrt[5]{\frac{1}{\sqrt{x}}+1}\right)^{9}+C=-\frac{10}{9}\left(\sqrt[5]{\frac{1+\sqrt{x}}{\sqrt{x}}}\right)^{9}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-17$ »

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

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## Problem Kuznetsov Integrals 13-18

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