## Problem Statement

Find the indefinite integral:

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

Let's represent the integral as a differential binomial:

$$
\int \frac{\sqrt[3]{\left(1+\sqrt[4]{x^{3}}\right)^{2}}}{x^{2} \cdot \sqrt[4]{x}} d x=\int x^{-9 / 4}\left(1+x^{3 / 4}\right)^{2 / 3} d x
$$

Perform a change of variable:

$$
\begin{aligned}
& x=\left(t^{3 / 2}-1\right)^{-4 / 3} \Rightarrow t=\left(x^{-3 / 4}+1\right)^{2 / 3} \\
& d x=-2 t^{1 / 2}\left(t^{3 / 2}-1\right)^{-2 / 3} d t
\end{aligned}
$$

Transition to the new variable of integration:

$$
\begin{aligned}
& \int x^{-9 / 4}\left(1+x^{3 / 4}\right)^{2 / 3} d x= \\
& \quad=-2 \int\left(\left(t^{3 / 2}-1\right)^{-4 / 3}\right)^{-9 / 4}\left(1+\left(\left(t^{3 / 2}-1\right)^{-4 / 3}\right)^{3 / 4}\right)^{2 / 3} t^{1 / 2}\left(t^{3 / 2}-1\right)^{-2 / 3} d t= \\
& \quad=-2 \int\left(t^{3 / 2}-1\right)^{3} t^{1 / 2}\left(1+\left(t^{3 / 2}-1\right)^{-1}\right)^{2 / 3}\left(t^{3 / 2}-1\right)^{-2 / 3} d t=
\end{aligned}
$$

Simplify and evaluate the resulting integral:

$$
=-2 \int t^{3 / 2} d t=-\frac{4}{5} t^{5 / 2}+c=
$$

Return to the original variable of integration:

$$
=-\frac{4}{5}\left(\left(x^{-3 / 4}+1\right)^{2 / 3}\right)^{5 / 2}+c=-\frac{4}{5}\left(x^{-3 / 4}+1\right)^{5 / 3}+c
$$

Answer:
$\int \frac{\sqrt[3]{\left(1+\sqrt[4]{x^{3}}\right)^{2}}}{x^{2} \cdot \sqrt[4]{x}} d x=-\frac{4}{5}\left(x^{-3 / 4}+1\right)^{5 / 3}+c$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_13-16" Categories: Kuznetsov's Problem Book Integrals Problem 13 | Integrals

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## Problem Kuznetsov Integrals 13-17

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