bemath-dx-4-3-2x-1-3-1-4-

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

\sqrt{bemath}

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

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

b y l e t t i n g ν = \sqrt[4]{4 - \sqrt[3]{3 - 2 x}}

\Rightarrow ν^{4} = 4 - \sqrt[3]{3 - 2 x} \Rightarrow 3 - 2 x = {(4 - ν^{4})}^{3}

2 x = 3 - {(4 - ν^{4})}^{3}

\Rightarrow 2 d x = - 3 . (- 4 ν^{3}) {(4 - ν^{4})}^{2} d ν

I = \int \frac{12 ν^{3} {(4 - ν^{4})}^{2}}{2 ν} d ν

I = 6 \int ν^{2} (16 - 8 ν^{4} + ν^{8}) d ν

I = 6 \int (16 ν^{2} - 8 ν^{6} + ν^{10}) d ν

I = 6 (\frac{16 ν^{3}}{3} - \frac{8}{7} ν^{7} + \frac{1}{11} ν^{11}) + c

I = 6 \sqrt[4]{{(4 - \sqrt[3]{3 - 2 x})}^{3}} (\frac{16}{3} - \frac{8}{7} (4 - \sqrt[3]{3 - 2 x}) + \frac{1}{11} {(4 - \sqrt[3]{3 - 2 x})}^{2}) + c

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

\int \frac{d x}{\sqrt[4]{4 - a}} 3 - 2 x = a^{3} - 2 = 3 a^{2} \frac{d a}{d x}

- \frac{1}{2} \int \frac{3 a^{2} d a}{{(4 - a)}^{\frac{1}{4}}} (4 - a = t^{4} \Rightarrow - 1 = 4 t^{3} \frac{d t}{d a}

\frac{3}{2} \int \frac{a^{2} 4 t^{3} d t}{t} = 6 \int t^{2} {(4 - t^{4})}^{2} d t = 6 \int 16 t^{2} - 8 t^{6} + t^{10}

= 32 t^{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 - 2 x})}^{\frac{3}{4}} - \frac{48}{7} {(4 - \sqrt[3]{3 - 2 x})}^{\frac{7}{4}} + \frac{6}{11} {(4 - \sqrt[3]{3 - 2 x})}^{\frac{11}{4}} + C