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find-dx-x-1-x-2-x-3-




Question Number 65355 by mathmax by abdo last updated on 28/Jul/19
find ∫   (dx/( (√((x+1)(x+2)(x+3)))))
finddx(x+1)(x+2)(x+3)
Commented by Prithwish sen last updated on 29/Jul/19
∫(dx/( (√({(x+2)^2 −1}(x+2)))))   putting (x+2)=a   =∫(da/( (√(a(a^2 −1)))))  again putting a=secθ  = ∫(√(secθ)) dθ  ????
dx{(x+2)21}(x+2)putting(x+2)=a=daa(a21)againputtinga=secθ=secθdθ????
Answered by ajfour last updated on 29/Jul/19
x+2=t  ∫(dx/( (√(t(t^2 −1)))))  let t=sec^2 θ ⇒  dt=2sec^2 θtan θdθ  ∫((2sec θtan θdθ)/( (√(sec^4 θ−1))))  let  sec θ=z  ∫((2dz)/( (√(z^4 −1))))= ?
x+2=tdxt(t21)lett=sec2θdt=2sec2θtanθdθ2secθtanθdθsec4θ1letsecθ=z2dzz41=?
Answered by MJS last updated on 29/Jul/19
∫(dx/( (√((x+1)(x+2)(x+3)))))=       [t=arccosh (x+2) → dx=dt(√((x+1)(x+3)))]  =∫(dt/( (√(cosh t))))=       [u=(t/2)i → dt=−2idu]  =−2i∫(du/( (√(cos 2u))))=−2i∫(du/( (√(1−2sin^2  u))))=  this is an incomplete elliptic integral  =−2iF(u∣2)  ...
dx(x+1)(x+2)(x+3)=[t=arccosh(x+2)dx=dt(x+1)(x+3)]=dtcosht=[u=t2idt=2idu]=2iducos2u=2idu12sin2u=thisisanincompleteellipticintegral=2iF(u2)

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