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Question Number 159612 by cortano last updated on 19/Nov/21

Answered by Ar Brandon last updated on 19/Nov/21

$$I=\int \frac{\sqrt{2+\sqrt[3]{x}}}{\sqrt[3]{x}}dx,x=u^3\Rightarrow dx=3u^{2}du$$$$=3\int \frac{\sqrt{2+u}}{u}\cdot u^{2}du=3\int u\sqrt{2+u}du$$$$=3\int {(u+2)}^{\frac{3}{2}}-2\sqrt{u+2}du$$$$=\frac{6}{5}{(u+2)}^{\frac{5}{2}}-4{(u+2)}^{\frac{3}{2}}+C$$$$=\frac{6}{5}{(\sqrt[3]{x}+2)}^{\frac{5}{2}}-4{(\sqrt[3]{x}+2)}^{\frac{3}{2}}+C$$

Answered by tounghoungko last updated on 19/Nov/21

$$I=\int \frac{\sqrt{2+\sqrt[3]{x}}}{\sqrt[3]{x}}dx;let\sqrt{2+\sqrt[3]{x}}=u$$$$\sqrt[3]{x}=u^2-2\Rightarrow \frac{dx}{\sqrt[3]{x}}=6u\sqrt[3]{u}du$$$$\frac{dx}{\sqrt[3]{x}}=6u(u^2-2)du=(6u^3-12u)du$$$$I=\int u(6u^3-12u)du$$$$I=\int (6u^4-12u^2)du$$$$I=\frac{6}{5}{u}^5-4u^3+c$$$$I=\frac{6}{5}{\left(\sqrt{2+\sqrt[3]{x}}\right)}^5-4{\left(\sqrt{2+\sqrt[3]{x}}\right)}^3+c$$

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