## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[3]{1+\sqrt[3]{x}}}{x \sqrt[5]{x^{2}}} d x
$$
## Solution

Transform the integrand:

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

This integral represents a differential binomial.

Since $\frac{\frac{-7}{5}+1}{\frac{1}{3}}+\frac{1}{5}=-1-$ is an integer, we make the substitution:

$$
x^{\frac{-1}{3}}+1=t^{5}
$$

Then

$$
\frac{-1}{3} x^{\frac{-4}{3}} d x=5 t^{4} d t, d x=-3 x^{\frac{4}{3}} 5 t^{4} d t
$$

We obtain:

$$
\begin{aligned}
& \int x^{-\frac{7}{5}}\left(1+x^{\frac{1}{3}}\right)^{\frac{1}{5}} d x=\int x^{-\frac{7}{5}}\left(x^{\frac{1}{3}}\left(x^{\frac{-1}{3}}+1\right)\right)^{\frac{1}{5}} d x=\int x^{-\frac{7}{5}} x^{\frac{1}{15}}\left(x^{\frac{-1}{3}}+1\right)^{\frac{1}{5}} d x= \\
& =\int x^{-\frac{4}{3}}\left(x^{\frac{-1}{3}}+1\right)^{\frac{1}{5}} d x=\int x^{-\frac{4}{3}}\left(t^{5}\right)^{\frac{1}{5}}\left(-3 x^{\frac{4}{3}} 5 t^{4}\right) d t=-15 \int t^{5} d t= \\
& =\frac{-15}{6} \cdot t^{6}+C=\frac{-5}{2}\left(x^{\frac{-1}{3}}+1\right)^{\frac{6}{5}}+C=\frac{-5}{2} \sqrt[3]{\left(x^{\frac{-1}{3}}+1\right)^{6}}+C= \\
& =-\frac{5}{2} \sqrt[5]{\frac{1+\sqrt[3]{x}}{x^{2}}}+C, \text { where } C-\text { is an arbitrary constant. }
\end{aligned}
$$

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 \%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB\%D1\%8B_13-31»

Categories: Kuznetsov's Integral Problems 13 | Integrals

- Last modified: 15:21, 29 May 2010.
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