如果把减号改成加号，就好办了。
=∫(ln(x+2)-ln(x+1))/(x+1)-(ln(x+2)-ln(x+1))/(x+2)dx=∫ln(x+2)dln(x+1)+∫ln(x+1)dln(x+2)-∫ln(x+1)dln(x+1)-∫ln(x+2)dln(x+2)=ln(x+1)ln(x+2)-ln²(x+1)/2-ln²(x+2)/2+C
