設dy/dx=(x^0.5)/[(6+x)^0.5 -x^0.5]，求y?

dy/dx = √x/[√(6+x)-√x] = √x[√(6+x)+√x] /6 = [√(6x+x^2)+x]/6
          = √[(x+3)^2-9]/6 + x/6,  x ≧ 0

∴ y =  (1/6)∫√[(x+3)^2-9] dx + x^2/12

令 x+3 = 3sec(t), 則 dx = 3sec(t)tan(t) dt
∴ ∫√[(x+3)^2-9] dx = ∫|3tan(t)|(3sec(t)tan(t)) dt
因為 x ≧ 0, x+3 = 3sec(t) 對應 0 ≦ t < π/2,
故前列積分式中 |tan(t)| = tan(t).
∴ ∫√[(x+3)^2-9] dx = ∫ 9sec(t)tan^2(t) dt
       
 ∫ 9sec(t)tan^2(t) dt = ∫ sec^3(t)-sec(t) dt 
    = ∫ sec^3(t) dt - ∫ sec(t) dt
    = sec(t)tan(t) - ∫ tan(t) sec(t)tan(t) dt - ∫ sec(t) dt
     = sec(t)tan(t) - ∫ sec(t)tan^2(t) dt - ∫ sec(t) dt
 
∴ ∫ sec(t)tan^2(t) dt 
        = (1/2){ sec(t)tan(t) - ∫ sec(t) dt }

∴  ∫√[(x+3)^2-9] dx 
         = (9/2) sec(t)tan(t) - (9/2) ∫ sec(t) dt
         = (9/2) sec(t)tan(t) - (9/2) ln|sec(t)+tan(t)| + C

sec(t) = (x+3)/3, 
tan(t) = √(sec^2(t)-1) = √[(x+3)^2-9]/3

∴ y = x^2/12+(1/6) ∫√[(x+3)^2-9] dx
        = x^2/12 + (9/12) [(x+3)/3]√[(x+3)^2-9]/3
                  -(9/12) ln|(x+3)/3+√[(x+3)^2-9]/3| + C
        = x^2/12 + (1/12)(x+3)√[(x+3)^2-9]
                  - (3/4) ln{(x+3)+√[(x+3)^2-9]} + C'