∑(n=1~∞) [(x+1)^n]/ [n(2^n)]
=∑(n=1~∞) ∫[(x+1)^(n-1)]/(2^n)dx
=1/2∫∑(n=1~∞) [(x+1)/2]^(n-1)]dx
=1/2∫[1/(1-(x+1)/2)]dx
=∫[1/(1-x)]dx (-3
∑(n=1~∞) [(x+1)^n]/ [n(2^n)]
=∑(n=1~∞) ∫[(x+1)^(n-1)]/(2^n)dx
=1/2∫∑(n=1~∞) [(x+1)/2]^(n-1)]dx
=1/2∫[1/(1-(x+1)/2)]dx
=∫[1/(1-x)]dx (-3