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

Question

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

My attempt requires two substitutions, I was wondering if there's any other way of solving this, including hopefully an elegant trig sub

let $x=u^4$ so we now have

$$\int 4u^3 \sqrt[3]{1+u } du $$

now , again let $1+u= z^3$

so we have $$12\int z^2(z)(z^3-1)^3dz$$

Which is

$12(\frac{z^{13}}{13} - \frac{3z^{10}}{10}+\frac{3z^7}{7} -\frac{z^4}{4} +1)$

which is an elementary , albeit lengthy integral , however, as mentioned earlier, I was wondering if there's any other way of solving this

Source :- A problem book in mathematical analysis

Answer 1

Let, $$ \sqrt[3]{1+ \sqrt[4]x}=u $$ and then $$ x=(u^3-1)^4.$$ So, $$\begin{eqnarray} &&\int \sqrt[3]{1+ \sqrt[4]x}dx=\int u\cdot4(u^3-1)^3\cdot3u^2du\\ &=&12\int u^3(u^3-1)^3du=\int(u^{12}-3u^9+3u^6-u^3)du\\ &=&\frac{12}{13}u^{13}-\frac{18}{5}u^{10}+\frac{36}7u^7-3u^4+C\\ &=&\frac{12}{13}\left(\sqrt[3]{1+ \sqrt[4]x}\right)^{13}-\frac{18}{5}\left(\sqrt[3]{1+ \sqrt[4]x}\right)^{10}+\frac{36}7\left(\sqrt[3]{1+ \sqrt[4]x}\right)^7-3\left(\sqrt[3]{1+ \sqrt[4]x}\right)^4+C. \end{eqnarray}$$

Answer 2

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

let $x=u^4$ So we now have

$$\int 4u^3 \sqrt[3]{1+u } du $$

now , again let $1+u= z^3$

so we have $$4\int (z^3-1)^3 (z) 3z^2dz$$

$$=12\int z^3(z^3-1)^3 dz$$ $$=12\int z^3(z^9-3z^6+3z^3-1) dz$$ $$=12\int (z^{12}-3z^9+3z^6-z^3) dz$$ $$=\frac{12z^{13}}{13}-\frac{36z^{10}}{10}+\frac{36z^{7}}{7}-\frac{12z^{4}}{4}+c$$ $$=\frac{12z^{13}}{13}-\frac{18z^{10}}{5}+\frac{36z^{7}}{7}-3z^{4}+c$$

$$=\frac{12}{13}\left(\sqrt[3]{1+ \sqrt[4]x}\right)^{13}-\frac{18}{5}\left(\sqrt[3]{1+ \sqrt[4]x}\right)^{10}+\frac{36}7\left(\sqrt[3]{1+ \sqrt[4]x}\right)^7-3\left(\sqrt[3]{1+ \sqrt[4]x}\right)^4+c$$

where $c$ is an arbitrary constant.

I think it is correct form of your method.It is not another method.