## Problem Statement

Find the indefinite integral:

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

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

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

$$
m=-2 ; n=\frac{3}{4} ; p=\frac{1}{3}
$$

Since $\frac{m+1}{n}+p=\frac{-2+1}{\frac{3}{4}}+\frac{1}{3}=-\frac{4}{3}+\frac{1}{3}=-1$ is an integer, we use the substitution:

$$
a x^{-n}+b=z^{s}, \text { where } s \text { is the denominator of the fraction } p .
$$

In our case, the substitution is:

$$
\begin{aligned}
& x^{-\frac{3}{4}}+1=z^{3} \\
& x=\left(z^{3}-1\right)^{-\frac{4}{3}} \\
& d x=-\frac{4}{3}\left(z^{3}-1\right)^{-\frac{7}{3}} \cdot 3 z^{2} d z=-4 z^{2}\left(z^{3}-1\right)^{-\frac{7}{3}} d z
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{3}-1\right)^{-\frac{4}{3}}\right)^{-2} \cdot\left(1+\left(\left(z^{3}-1\right)^{-\frac{4}{3}}\right)^{\frac{3}{4}}\right)^{\frac{1}{3}}(-4) z^{2}\left(z^{3}-1\right)^{-\frac{7}{3}} d z= \\
& =-4 \cdot \int\left(z^{3}-1\right)^{\frac{8}{3}} \cdot\left(1+\frac{1}{z^{3}-1}\right)^{\frac{1}{3}} z^{2}\left(z^{3}-1\right)^{-\frac{7}{3}} d z= \\
& =-4 \cdot \int\left(z^{3}-1\right)^{\frac{1}{3}} \cdot\left(\frac{z^{3}}{z^{3}-1}\right)^{\frac{1}{3}} z^{2} d z=-4 \cdot \int\left(z^{3}-1\right)^{\frac{1}{3}} \cdot \frac{z}{\left(z^{3}-1\right)^{\frac{1}{3}}} z^{2} d z=
\end{aligned}
$$

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-15 »$ Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last edited: 15:46, 19 July 2009.
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## Problem Kuznetsov Integrals 13-16

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