$\int \sqrt[3]{x}\left\{1+\sqrt[3]{x^4}\right\}^{\frac{1}{7}} d x=A\left(1+\sqrt[3]{x^4}\right)^B+c$
Let us assume
$$
I=\int \sqrt[3]{x}\left(1+\sqrt[3]{x^4}\right)^{\frac{1}{7}}=\int x^{\frac{1}{3}}\left(1+x^{\frac{4}{3}}\right)^{\frac{1}{7}} d x
$$
$\begin{aligned} & \text { Let } 1+x^{\frac{4}{3}}=t \\ & \begin{aligned} \Rightarrow \frac{4}{3} x^{\frac{4}{3}-1} d x & =d t \Rightarrow \frac{4}{3} x^{\frac{1}{3}} d x=d t \\ \Rightarrow \quad x^{\frac{1}{3}} d x & =\frac{3}{4} d t \\ \therefore \quad I & =\int(t)^{\frac{1}{7}} \frac{3}{4} d t=\frac{3}{4} \int(t)^{\frac{1}{7}} d t \\ & =\frac{3}{4}\left[\frac{(t)^{\frac{1}{7}}+1}{\frac{1}{7}+1}\right]+c=\frac{3}{4} \frac{\left(1+x^{4 / 3}\right)^{\frac{8}{7}}}{\frac{8}{7}}+c \\ & =\frac{21}{32}\left(1+\sqrt[3]{x^4}\right)^{\frac{8}{7}}+c\end{aligned}\end{aligned}$
$\begin{aligned} & \therefore A=\frac{21}{32} \text { and } B=\frac{8}{7} \\ & \therefore A B=\frac{21}{32} \times \frac{8}{7}=\frac{3}{4}\end{aligned}$