## Problem Statement

Find the indefinite integral:

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

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

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

$$
m=-\frac{5}{3} ; n=\frac{1}{2} ; p=\frac{1}{3}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{5}{3}+1}{\frac{1}{2}}+\frac{1}{3}=-\frac{4}{3}+\frac{1}{3}=-1$ 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:

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

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{3}-1\right)^{-2}\right)^{-\frac{5}{3}} \cdot\left(1+\left(\left(z^{3}-1\right)^{-2}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}}(-6) z^{2}\left(z^{3}-1\right)^{-3} d z= \\
& =-6 \cdot \int\left(z^{3}-1\right)^{\frac{10}{3}} \cdot\left(1+\frac{1}{z^{3}-1}\right)^{\frac{1}{3}} z^{2}\left(z^{3}-1\right)^{-3} d z= \\
& =-6 \cdot \int\left(z^{3}-1\right)^{\frac{1}{3}} \cdot\left(\frac{z^{3}}{z^{3}-1}\right)^{\frac{1}{3}} z^{2} d z=-6 \cdot \int\left(z^{3}-1\right)^{\frac{1}{3}} \cdot \frac{z}{\left(z^{3}-1\right)^{\frac{1}{3}}} z^{2} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-6 \cdot \int z^{3} d z=-6 \cdot \frac{z^{4}}{4}+C=-\frac{3 z^{4}}{2}+C=-\frac{3\left(\sqrt[3]{x^{-\frac{1}{2}}+1}\right)^{4}}{2}+C= \\
& =-\frac{3\left(\frac{1}{\sqrt{x}}+1\right)^{\frac{4}{3}}}{2}+C=-\frac{3(1+\sqrt{x})^{\frac{4}{3}}}{2 x^{\frac{2}{3}}}+C=-\frac{3 \sqrt[3]{(1+\sqrt{x})^{4}}}{2 \sqrt[3]{x^{2}}}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-2$ »

Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

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## Problem Kuznetsov Integrals 13-3

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