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solve-dx-c-b-ax-1-3-




Question Number 110245 by Her_Majesty last updated on 28/Aug/20
solve ∫(dx/( ((c−(√(b−ax))))^(1/3) ))
solvedxcbax3
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
dxcbax3Solutionsetu=baxdu=adx1aducu3sety=udu=2ydy2aydycy3setcy=wdw=dy2acww3dw2caw23×322aw113+1×353caw2365aw53+C
Commented by Her_Majesty last updated on 28/Aug/20
thank you
thankyou
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
inonestept=cbax3dx=6t2(t3c)a6a(t4ct)dt=3a(ct22t55)==35a(cbax)2/3(3c+2bax)+C
Commented by bobhans last updated on 28/Aug/20
cooll....great...nice
cooll.greatnice
Commented by bobhans last updated on 28/Aug/20
but what is “s“
butwhatiss
Commented by Her_Majesty last updated on 28/Aug/20
typo s=a
typos=a

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