∴ Original formula = 3 ∫ (0, 1) dx ∫ (0,1) ln [(y+1)] dy. And, ∫ (0,1) ln [(y+1)] dy = 2ln2-1-ln (1+x),

∴ Original formula = 3 ∫ (0,1) [2LN2-1-ln (1+x)] dx = 0.

For reference.