If int sqrt[3]{x}left{1+sqrt[3]{x^4}right}^{1 | vedclass

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(B) Let $I = \int \sqrt[3]{x}\left(1+\sqrt[3]{x^4}\right)^{\frac{1}{7}} d x = \int x^{\frac{1}{3}}\left(1+x^{\frac{4}{3}}\right)^{\frac{1}{7}} d x$. L

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$3 / 4$

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(B) Let $I = \int \sqrt[3]{x}\left(1+\sqrt[3]{x^4}ight)^{\frac{1}{7}} d x = \int x^{\frac{1}{3}}\left(1+x^{\frac{4}{3}}ight)^{\frac{1}{7}} d x$. Let $1+x^{\frac{4}{3}} = t$. Then,$\frac{4}{3} x^{\frac{1}{3}} d x = d t$,which implies $x^{\frac{1}{3}} d x = \frac{3}{4} d t$. Substituting these into the integral,we get: $I = \int t^{\frac{1}{7}} \cdot \frac{3}{4} d t = \frac{3}{4} \int t^{\frac{1}{7}} d t$. Integrating,we have: $I = \frac{3}{4} \cdot \frac{t^{\frac{1}{7}+1}}{\frac{1}{7}+1} + c = \frac{3}{4} \cdot \frac{t^{\frac{8}{7}}}{\frac{8}{7}} + c = \frac{3}{4} \cdot \frac{7}{8} t^{\frac{8}{7}} + c = \frac{21}{32} \left(1+x^{\frac{4}{3}}ight)^{\frac{8}{7}} + c$. Comparing this with $A\left(1+\sqrt[3]{x^4}ight)^B+c$,we get $A = \frac{21}{32}$ and $B = \frac{8}{7}$. Therefore,$A B = \frac{21}{32} 	imes \frac{8}{7} = \frac{3}{4}$.

If $\int \sqrt[3]{x}\left\{1+\sqrt[3]{x^4}\right\}^{1 / 7} d x=A\left(1+\sqrt[3]{x^4}\right)^B+c$,then the value of $A B$ is equal to

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