Triple integral 1/(x+y+z+1)³

Reading. The integrand is written with dx dy dz as the differentials, but the limits are for z innermost (0 to 1−x−y), then y (0 to 1−x), then x (0 to 1). I integrate z first, then y, then x.

I = ∫₀¹ ∫₀^{1−x} ∫₀^{1−x−y} dz dy dx / (x + y + z + 1)³

z-integration

∫₀^{1−x−y} (x+y+z+1)⁻³ dz = [ −1/(2(x+y+z+1)²) ]₀^{1−x−y}
                          = (1/2) [ 1/(x+y+1)² − 1/4 ]

(At z = 1 − x − y, x + y + z + 1 = 2.)

y-integration

∫₀^{1−x} (1/2)[ (x+y+1)⁻² − 1/4 ] dy
 = (1/2) [ −1/(x+y+1) − y/4 ]₀^{1−x}
 = (1/2) [ −1/2 − (1−x)/4 + 1/(x+1) ]

x-integration

I = (1/2) ∫₀¹ [ 1/(1+x) − 1/2 − (1−x)/4 ] dx
  = (1/2) [ ln 2 − 1/2 − 1/8 ]
  = (1/2) ln 2 − 5/16

Answer: I = (1/2) ln 2 − 5/16 ≈ 0.0341.