Funny math problem to calculate the Pin Number for a credit card.

[Albert Chan]

(08-12-2021 03:05 PM)Albert Chan Wrote:  I = ∫((3y^3 + 25y^2 + 77y + 55) / √(y^2-1), y=-3 .. -1) / 8

Another way, by regroup terms. (odd and even powers of y)

diff(y*√(y^2-1), y) = (2*y^2-1) / √(y^2-1))

I = ∫((3/8*(y^2-1)+10)*y/√(y^2-1), y=-3 .. -1) + ∫((25/16*(2*y^2-1)+135/16)/√(y^2-1), y=-3 .. -1)

For first term, let z = y^2-1, dz = 2y dy:

∫((3/8*(y^2-1)+10)*y/√(y^2-1), y) = ∫(3/16*√z+5/√z, z) = 1/8*z^(3/2) + 10*√z

I = preval((z/8+10)*√z,8,0,z) + preval(25/16*(y*√(y^2-1)) + 135/16*(-ln(abs(-y+√(y^2-1)))),-3,-1,y)
  = -22*√(2) + (150*√(2)+135*ln(2*√(2)+3))/16
  = -101/8*√(2) + 135/16*ln(2*√(2)+3)