## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[3]{1+\sqrt[5]{x}}}{x \sqrt[15]{x^{4}}} d x
$$
## Solution

$$
\int \frac{\sqrt[3]{1+\sqrt[5]{x}}}{x \sqrt[15]{x^{4}}} d x=\int x^{-\frac{19}{15}}\left(1+x^{\frac{1}{5}}\right)^{\frac{1}{3}} d x=
$$

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

$$
m=-\frac{19}{15} ; n=\frac{1}{5} ; p=\frac{1}{3}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{19}{15}+1}{\frac{1}{5}}+\frac{1}{3}=-\frac{4}{3}+\frac{1}{3}=-1$ 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:

$$
\begin{aligned}
& x^{-\frac{1}{5}}+1=z^{3} \\
& x=\left(z^{3}-1\right)^{-5} \\
& d x=-5\left(z^{3}-1\right)^{-6} \cdot 3 z^{2} d z=-15 z^{2}\left(z^{3}-1\right)^{-6} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{3}-1\right)^{-5}\right)^{-\frac{19}{15}} \cdot\left(1+\left(\left(z^{3}-1\right)^{-5}\right)^{\frac{1}{5}}\right)^{\frac{1}{3}} \cdot(-15) z^{2}\left(z^{3}-1\right)^{-6} d z= \\
& =-15 \cdot \int\left(z^{3}-1\right)^{\frac{19}{3}} \cdot\left(1+\frac{1}{z^{3}-1}\right)^{\frac{1}{3}} \cdot z^{2}\left(z^{3}-1\right)^{-\frac{18}{3}} d z= \\
& =-15 \cdot \int\left(z^{3}-1\right)^{\frac{1}{3}} \cdot\left(\frac{z^{3}}{z^{3}-1}\right)^{\frac{1}{3}} \cdot z^{2} \cdot d z=-15 \cdot \int z^{3} d z=
\end{aligned}
$$

$$
\begin{aligned}
& \left.=-15 \cdot \frac{z^{4}}{4}+C=-\frac{15 z^{4}}{4}+C=-\frac{15\left(\sqrt[3]{x^{-\frac{1}{5}}}+1\right.}{4}\right)^{4}+C= \\
& =-\frac{15}{4} \cdot\left(\sqrt[3]{\frac{1}{\sqrt[5]{x}}+1}\right)^{4}+C=-\frac{15}{4} \cdot\left(\sqrt[3]{\frac{1+\sqrt[5]{x}}{\sqrt[5]{x}}}\right)^{4}+C=-\frac{15(1+\sqrt[5]{x})^{4 / 3}}{4 x^{4 / 15}}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-30$ » Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last edited: 08:50, 24 May 2010.
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## Problem Kuznetsov Integrals 13-31

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