(12C) log1p function

[Albert Chan]

Dieter's asinh formula look very nice: log1p(x) = asinh ( (x²+2x)/(2x+2) ) = asinh(x - x/(2+2/x))

We can create "half angle" version, using symmetrical relation: atanh(sin(z)) = asinh(tan(z))

log1p(x) = 2 atanh( x/(x+2) ) = 2 asinh( x/ √((x+2)² - x²) )

log1p(x) = 2 asinh ( x/√(4x+4) )

Comment: it may be easier to derive relation with Arc SOHCAHTOA method, for hyperbolics

atanhq(x²/(x+2)²)      // TOA, O=x², A=(x+2)²
= asinhq(x²/(4x+4))   // SOH, H = A-O = 4x+4