∫㏑(1+x)dx
=∫㏑(1+x)d(x+1)
=(x+1)ln(x+1)-∫(x+1)d㏑(1+x)
=(x+1)ln(x+1)-∫(x+1)*1/(x+1)dx
=(x+1)ln(x+1)-∫dx
=(x+1)ln(x+1)-x+C