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Question Number 111017 by bemath last updated on 01/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\int \frac{\mathrm{dx}}{\sqrt[4]{4-\sqrt[3]{3-2\mathrm{x}}}}?$$

Answered by john santu last updated on 01/Sep/20

$$byletting\nu =\sqrt[4]{4-\sqrt[3]{3-2x}}$$$$\Rightarrow {\nu }^4=4-\sqrt[3]{3-2x}\Rightarrow 3-2x={(4-{\nu }^4)}^3$$$$2x=3-{(4-{\nu }^4)}^3$$$$\Rightarrow 2dx=-3.(-4{\nu }^3){(4-{\nu }^4)}^{2}d\nu$$$$I=\int \frac{12{\nu }^3{(4-{\nu }^4)}^2}{2\nu }d\nu$$$$I=6\int {\nu }^2(16-8{\nu }^4+{\nu }^8)d\nu$$$$I=6\int (16{\nu }^2-8{\nu }^6+{\nu }^{10})d\nu$$$$I=6(\frac{16{\nu }^3}{3}-\frac{8}{7}{\nu }^7+\frac{1}{11}{\nu }^{11})+c$$$$I=6\sqrt[4]{{(4-\sqrt[3]{3-2x})}^3}(\frac{16}{3}-\frac{8}{7}(4-\sqrt[3]{3-2x})+\frac{1}{11}{(4-\sqrt[3]{3-2x})}^2)+c$$

Answered by Dwaipayan Shikari last updated on 01/Sep/20

$$\int \frac{dx}{\sqrt[4]{4-a}}3-2x=a^3-2=3a^2\frac{da}{dx}$$$$-\frac{1}{2}\int \frac{3a^{2}da}{{(4-a)}^{\frac{1}{4}}}(4-a=t^4\Rightarrow -1=4t^3\frac{dt}{da}$$$$\frac{3}{2}\int \frac{a^{2}4t^{3}dt}{t}=6\int t^2{(4-t^4)}^{2}dt=6\int 16t^2-8t^6+t^{10}$$$$=32t^3-\frac{48}{7}{t}^7+\frac{6}{11}{t}^{11}+C$$$$=32{(4-a)}^{\frac{3}{4}}-\frac{48}{7}{(4-a)}^{\frac{7}{4}}+\frac{6}{11}{(4-a)}^{\frac{11}{4}}+C$$$$=32{(4-\sqrt[3]{3-2x})}^{\frac{3}{4}}-\frac{48}{7}{(4-\sqrt[3]{3-2x})}^{\frac{7}{4}}+\frac{6}{11}{(4-\sqrt[3]{3-2x})}^{\frac{11}{4}}+C$$

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