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Question Number 110245 by Her_Majesty last updated on 28/Aug/20

solve ∫(dx/( ((c−(√(b−ax))))^(1/3) ))

$${solve}\:\int\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{c}−\sqrt{{b}−{ax}}}} \\ $$

Commented by Lordose last updated on 28/Aug/20

∫(dx/((c−(√(b−ax))))^(1/3) ) ★Solution set u=b−ax du=−adx −(1/a)∫(du/((c−(√u)))^(1/3) ) set y=(√u) du=2ydy ((−2)/a)∫((ydy)/((c−y))^(1/3) ) set c−y=w dw=−dy (2/a)∫((c−w)/(w)^(1/3) )dw ((2c)/a)w^(2/3) ×(3/2)−(2/a)w^(1−(1/3)+1) ×(3/5) ((3c)/a)w^(2/3) −(6/(5a))w^(5/3) +C

$$\int\frac{\boldsymbol{\mathrm{dx}}}{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{c}}−\sqrt{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{ax}}}}} \\ $$$$\bigstar\boldsymbol{\mathrm{Solution}} \\ $$$$\boldsymbol{\mathrm{set}}\:\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{ax}} \\ $$$$\boldsymbol{\mathrm{du}}=−\boldsymbol{\mathrm{adx}} \\ $$$$−\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\int\frac{\boldsymbol{\mathrm{du}}}{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{c}}−\sqrt{\boldsymbol{\mathrm{u}}}}} \\ $$$$\boldsymbol{\mathrm{set}}\:\boldsymbol{\mathrm{y}}=\sqrt{\boldsymbol{\mathrm{u}}} \\ $$$$\boldsymbol{\mathrm{du}}=\mathrm{2}\boldsymbol{\mathrm{ydy}} \\ $$$$\frac{−\mathrm{2}}{\boldsymbol{\mathrm{a}}}\int\frac{\boldsymbol{\mathrm{ydy}}}{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{y}}}} \\ $$$$\boldsymbol{\mathrm{set}}\:\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{w}} \\ $$$$\boldsymbol{\mathrm{dw}}=−\boldsymbol{\mathrm{dy}} \\ $$$$\frac{\mathrm{2}}{\boldsymbol{\mathrm{a}}}\int\frac{\boldsymbol{\mathrm{c}}−\boldsymbol{\mathrm{w}}}{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{w}}}}\boldsymbol{\mathrm{dw}} \\ $$$$\frac{\mathrm{2}\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{a}}}\boldsymbol{\mathrm{w}}^{\frac{\mathrm{2}}{\mathrm{3}}} ×\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{2}}{\boldsymbol{\mathrm{a}}}\boldsymbol{\mathrm{w}}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}} ×\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{\mathrm{3}\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{a}}}\boldsymbol{\mathrm{w}}^{\frac{\mathrm{2}}{\mathrm{3}}} −\frac{\mathrm{6}}{\mathrm{5}\boldsymbol{\mathrm{a}}}\boldsymbol{\mathrm{w}}^{\frac{\mathrm{5}}{\mathrm{3}}} +\boldsymbol{\mathrm{C}} \\ $$$$ \\ $$

Commented by Her_Majesty last updated on 28/Aug/20

thank you

$${thank}\:{you} \\ $$

Answered by Her_Majesty last updated on 28/Aug/20

in one step t=((c−(√(b−ax))))^(1/3) ⇒ dx=−((6t^2 (t^3 −c))/a) −(6/a)∫(t^4 −ct)dt=(3/a)(ct^2 −((2t^5 )/5))= =(3/(5a))(c−(√(b−ax)))^(2/3) (3c+2(√(b−ax)))+C

$${in}\:{one}\:{step} \\ $$$${t}=\sqrt[{\mathrm{3}}]{{c}−\sqrt{{b}−{ax}}}\:\Rightarrow\:{dx}=−\frac{\mathrm{6}{t}^{\mathrm{2}} \left({t}^{\mathrm{3}} −{c}\right)}{{a}} \\ $$$$−\frac{\mathrm{6}}{{a}}\int\left({t}^{\mathrm{4}} −{ct}\right){dt}=\frac{\mathrm{3}}{{a}}\left({ct}^{\mathrm{2}} −\frac{\mathrm{2}{t}^{\mathrm{5}} }{\mathrm{5}}\right)= \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}{a}}\left({c}−\sqrt{{b}−{ax}}\right)^{\mathrm{2}/\mathrm{3}} \left(\mathrm{3}{c}+\mathrm{2}\sqrt{{b}−{ax}}\right)+{C} \\ $$

Commented by bobhans last updated on 28/Aug/20

cooll....great...nice

$${cooll}....{great}...{nice} \\ $$

Commented by bobhans last updated on 28/Aug/20

but what is “s“

$${but}\:{what}\:{is}\:``{s}`` \\ $$

Commented by Her_Majesty last updated on 28/Aug/20

typo s=a

$${typo}\:{s}={a} \\ $$

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