## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{\sqrt[3]{(1+\sqrt[4]{x})^{2}}}{x \sqrt[12]{x^{5}}} d x
$$
## Solution

$$
\int \frac{\sqrt[3]{(1+\sqrt[4]{x})^{2}}}{x \sqrt[12]{x^{5}}} d x=\int x^{-\frac{17}{12}}\left(1+x^{\frac{1}{4}}\right)^{\frac{2}{3}} d x=
$$

By Chebyshev's substitutions, this is the 3rd case. Therefore, we get:

$$
\begin{aligned}
& \int \frac{\sqrt[3]{(1+\sqrt[4]{x})^{2}}}{x \sqrt[12]{x^{5}}} d x=\left|1+\sqrt[4]{x}=\sqrt[4]{x} \cdot t^{3} ; t=\sqrt[3]{\frac{1+\sqrt[4]{x}}{\sqrt[4]{x}}} ; x=\frac{1}{\left(t^{3}-1\right)^{2}} ; d x=-\frac{12 t^{2} d t}{\left(t^{3}-1\right)^{5}}\right|= \\
& =-12 \int \frac{\sqrt[3]{\left(1+\frac{1}{t^{3}-1}\right)^{2}} \cdot t^{2} d t}{\frac{\left(t^{3}-1\right)^{5}}{\left(t^{3}-1\right)^{4}} \cdot \sqrt[12]{\frac{1}{\left(t^{3}-1\right)^{20}}}}=-12 \int \frac{\sqrt[3]{\left(\frac{t^{3}}{t^{3}-1}\right)^{2}} \cdot t^{2} d t}{\sqrt[\frac{1}{\left(t^{3}-1\right)^{2}}]{ }}=-12 \int t^{4} d t=-\frac{12}{5} t^{5}=-\frac{12}{5} \sqrt[3]{\left(\frac{1+\sqrt[4]{x}}{\sqrt[4]{x}}\right)^{5}}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_13-27$ »

Categories: Kuznetsov's Problem Book Integrals Problem $13 \mid$ Integrals

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## Problem Kuznetsov Integrals 13-28

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