## Problem Statement

Find the indefinite integral:

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

## Problem 13.

Find the indefinite integral:

$$
\int \frac{\sqrt[3]{1+\sqrt[4]{x}}}{x \sqrt[3]{x}} d x
$$

Represent the integral in the form of a differential binomial:

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

Perform the substitution of the variable:

$x=\left(t^{3}-1\right)^{-4} \Rightarrow t=\left(x^{-1 / 4}+1\right)^{1 / 3}$

$$
\mathrm{dx}=-12 \mathrm{t}^{2}\left(\mathrm{t}^{3}-1\right)^{-5} \mathrm{dt}
$$

Transition to the new variable of integration:

$$
\begin{aligned}
& \int x^{-4 / 3}\left(1+x^{1 / 4}\right)^{1 / 3} d x= \\
& =-12 \int\left(t^{3}-1\right)^{16 / 3} t^{2}\left(1+\left(t^{3}-1\right)^{-1}\right)^{1 / 3}\left(t^{3}-1\right)^{-5} d t
\end{aligned}
$$

Simplify and evaluate the resulting integral:

$$
\begin{aligned}
& -12 \int\left(t^{3}-1\right)^{16 / 3} t^{2}\left(1+\left(t^{3}-1\right)^{-1}\right)^{1 / 3}\left(t^{3}-1\right)^{-5} d t= \\
& =-12 \int t^{3} d t=-3 t^{4}+c
\end{aligned}
$$

Return to the original variable of integration:

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

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

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_13-26»

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## Problem Kuznetsov Integrals 13-27

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