## Problem Statement

Find the indefinite integral:

$$
\int \frac{\sqrt[4]{(1+\sqrt{x})^{3}}}{x \sqrt[8]{x^{7}}} d x
$$
## Solution

$$
\int \frac{\sqrt[4]{(1+\sqrt{x})^{3}}}{x \sqrt[8]{x^{7}}} d x=\int x^{-\frac{15}{8}}\left(1+x^{\frac{1}{2}}\right)^{\frac{3}{4}} d x=
$$

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

$$
m=-\frac{15}{8} ; n=\frac{1}{2} ; p=\frac{3}{4}
$$

Since $\frac{m+1}{n}+p=\frac{-\frac{15}{8}+1}{\frac{1}{2}}+\frac{3}{4}=-\frac{7}{4}+\frac{3}{4}=-1$ is an integer, we use the substitution:

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

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

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

We get:

$$
\begin{aligned}
& =\int\left(\left(z^{4}-1\right)^{-2}\right)^{-\frac{15}{8}} \cdot\left(1+\left(\left(z^{4}-1\right)^{-2}\right)^{\frac{1}{2}}\right)^{\frac{3}{4}}(-8) z^{3}\left(z^{4}-1\right)^{-3} d z= \\
& =-8 \int\left(z^{4}-1\right)^{\frac{15}{4}} \cdot\left(1+\frac{1}{z^{4}-1}\right)^{\frac{3}{4}} z^{3}\left(z^{4}-1\right)^{-3} d z= \\
& =-8 \int\left(z^{4}-1\right)^{\frac{3}{4}} \cdot\left(\frac{z^{4}}{z^{4}-1}\right)^{\frac{3}{4}} z^{3} d z=-8 \int\left(z^{4}-1\right)^{\frac{3}{4}} \cdot \frac{z^{3}}{\left(z^{4}-1\right)^{\frac{3}{4}}} z^{3} d z=
\end{aligned}
$$

$$
\begin{aligned}
& =-8 \int z^{6} d z=-8 \cdot \frac{z^{7}}{7}+C=\frac{-8\left(\sqrt[4]{x^{-\frac{1}{2}}+1}\right)^{7}}{7}+C= \\
& =-\frac{8}{7}\left(\sqrt[4]{\frac{1}{\sqrt{x}}+1}\right)^{7}+C=-\frac{8}{7}\left(\sqrt[4]{\frac{1+\sqrt{x}}{\sqrt{x}}}\right)^{7}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+13-11$ » Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

- Last edited: 12:09, 18 November 2009.
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## Problem Kuznetsov Integrals 13-12

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