## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt{1+\sqrt[4]{x^{3}}}}{x^{2} \cdot \sqrt[8]{x}} d x
$$
## Solution

$$
\int \frac{\sqrt{1+\sqrt[4]{x^{3}}}}{x^{2} \cdot \sqrt[8]{x}} d x=\int x^{-\frac{17}{8}}\left(1+x^{\frac{3}{4}}\right)^{\frac{1}{2}} d x=
$$

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

$$
m=-\frac{17}{8} ; n=\frac{3}{4} ; p=\frac{1}{2}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{17}{8}+1}{\frac{3}{4}}+\frac{1}{2}=-\frac{3}{2}+\frac{1}{2}=-1$ is an integer, we use the substitution:

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

That is, in our case, the substitution has the form:

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

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{2}-1\right)^{-\frac{4}{3}}\right)^{-\frac{17}{8}} \cdot\left(1+\left(\left(z^{2}-1\right)^{-\frac{4}{3}}\right)^{\frac{3}{4}}\right)^{\frac{1}{2}}\left(-\frac{8}{3}\right) z\left(z^{2}-1\right)^{-\frac{7}{3}} d z= \\
& =-\frac{8}{3} \cdot \int\left(z^{2}-1\right)^{\frac{17}{6}} \cdot\left(1+\frac{1}{z^{2}-1}\right)^{\frac{1}{2}} z\left(z^{2}-1\right)^{-\frac{14}{6}} d z= \\
& =-\frac{8}{3} \cdot \int\left(z^{2}-1\right)^{\frac{1}{2}} \cdot\left(\frac{z^{2}}{z^{2}-1}\right)^{\frac{1}{2}} z d z=-\frac{8}{3} \cdot \int\left(z^{2}-1\right)^{\frac{1}{2}} \cdot \frac{z}{\left(z^{2}-1\right)^{\frac{1}{2}}} z d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-\frac{8}{3} \cdot \int z^{2} d z=-\frac{8}{3} \cdot \frac{z^{3}}{3}+C=-\frac{8 z^{3}}{9}+C=-\frac{8\left(\sqrt{x^{-\frac{3}{4}}+1}\right)^{3}}{9}+C= \\
& =-\frac{8}{9} \sqrt{\left(\frac{1}{\sqrt[4]{x^{3}}}+1\right)^{3}}+C=-\frac{8}{9} \sqrt{\left(\frac{1+\sqrt[4]{x^{3}}}{\sqrt[4]{x^{3}}}\right)^{3}}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-14$ »

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

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## Problem Kuznetsov Integrals 13-15

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