If int frac{sqrt[4]{x}}{sqrt{x}+sqrt[4]{x}} d | vedclass

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(B) Let $I = \int \frac{\sqrt[4]{x}}{\sqrt{x}+\sqrt[4]{x}} dx$. Substitute $u = \sqrt[4]{x}$,then $x = u^4$ and $dx = 4u^3 du$. Substituting these int

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$-\frac{2}{3}$

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(B) Let $I = \int \frac{\sqrt[4]{x}}{\sqrt{x}+\sqrt[4]{x}} dx$. Substitute $u = \sqrt[4]{x}$,then $x = u^4$ and $dx = 4u^3 du$. Substituting these into the integral: $I = \int \frac{u}{u^2+u} (4u^3 du) = 4 \int \frac{u^4}{u(u+1)} du = 4 \int \frac{u^3}{u+1} du$. Using polynomial division,$\frac{u^3}{u+1} = u^2 - u + 1 - \frac{1}{u+1}$. $I = 4 \int (u^2 - u + 1 - \frac{1}{u+1}) du = 4 (\frac{u^3}{3} - \frac{u^2}{2} + u - \log|u+1|) + K$. $I = \frac{4}{3}u^3 - 2u^2 + 4u - 4 \log(u+1) + K$. Factoring out $\frac{2}{3}$: $I = \frac{2}{3} [2u^3 - 3u^2 + 6u - 6 \log(u+1)] + K$. Substituting $u = \sqrt[4]{x}$: $I = \frac{2}{3} [2 \sqrt[4]{x^3} - 3 \sqrt[4]{x^2} + 6 \sqrt[4]{x} - 6 \log(1+\sqrt[4]{x})] + K$. Comparing with the given form,we get $A=2, B=-3, C=6, D=-6$. Thus,$\frac{2}{3}(A+B+C+D) = \frac{2}{3}(2 - 3 + 6 - 6) = \frac{2}{3}(-1) = -\frac{2}{3}$.

If $\int \frac{\sqrt[4]{x}}{\sqrt{x}+\sqrt[4]{x}} d x=\frac{2}{3}\left[A \sqrt[4]{x^3}+B \sqrt[4]{x^2}+C \sqrt[4]{x}+D \log (1+\sqrt[4]{x})\right]+K$ then $\frac{2}{3}(A+B+C+D)=$

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