原式=∫ln(x+1)d(x+1) =(x+1)ln(x+1)-∫(x+1)dln(x+1) =(x+1)ln(x+1)-∫(x+1)*1/(x+1)dx =(x+1)ln(x+1)-∫dx =(x+1)ln(x+1)-x [0→1/e] =[(1/e+1)ln(1/e+1)-1/e]-[1*ln1-0] =(1/e+1)ln(1/e+1)-1/e
