How to integrate sqrt(1+x^(1/3))

 

$$\int 1/\surd (1+∛x)dx$$

By Substitution

$$\begin{array}{c}\text{ Suppose }\begin{array}{rl}\Rightarrow z& =\sqrt[3]{x}\\ z^3& =x\end{array}\\ 3z^{2}dz=dx\\ \int \frac{3z^2}{\sqrt{1+z}}\mathrm{d}z\end{array}$$ By Substitution

$$Suppose\Rightarrow u=\surd (1+z)$$

$$\begin{array}{l}{u}^2=1+z,z^2={(u^2-1)}^2\\ 2udu=dz\\ \int \frac{3{(u^2-1)}^2}{u}du\\ 3\int \frac{u^4-2u^2+1}{u}du\\ 3\int u^3-2u+\frac{1}{u}du\\ 3(\frac{u^4}{4}-u^2+\ln u)+c\\ 3(\frac{(\sqrt{1+z}{)}^4}{4}-(\sqrt{1+z}{)}^2+\ln |\sqrt{1+z}|)+c\\ 3(\frac{(\sqrt{1+\sqrt[3]{x}}{)}^4}{4}-(\sqrt{1+\sqrt[3]{x}}{)}^2+\ln |\sqrt{1+\sqrt[3]{x}}|)+c\end{array}$$