## Problem Statement

Find the indefinite integral:

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

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

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

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

Since $\frac{m+1}{n}+p=\frac{-2+1}{\frac{2}{3}}+\frac{1}{2}=-\frac{3}{2}+\frac{1}{2}=-1$ 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:

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

We get:

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

$$
\begin{aligned}
& =-3 \cdot \int z^{2} d z=-3 \cdot \frac{z^{3}}{3}+C=z^{3}+C=\left(\sqrt{x^{-\frac{2}{3}}+1}\right)^{3}+C= \\
& =\sqrt{\left(\frac{1}{\sqrt[3]{x^{2}}}+1\right)^{3}}+C=\sqrt{\left(\frac{1+\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}\right)^{3}}+C=\frac{\sqrt{\left(1+\sqrt[3]{x^{2}}\right)^{3}}}{x}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-9$ » Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last edited: 11:47, 21 July 2009.
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## Problem Kuznetsov Integrals 13-10

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