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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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