原式=1/2∫ln(x+1)dx²
=1/2*x²ln(x+1)-1/2∫x²dln(x+1)
=1/2*x²ln(x+1)-1/2∫x²/(x+1) dx
=1/2*x²ln(x+1)-1/2∫(x²-1+1)/(x+1) dx
=1/2*x²ln(x+1)-1/2∫[x-1+1/(x+1)] dx
=1/2*x²ln(x+1)-1/4*x²+1/2x-1/2ln(x+1)+C�����е������е�ת��û̫����������ϸ�Ľ����ô��Ϊʲô�ڶ���(x^2-1+1)����һ����ֱ�ӱ����(x-1+1)=(x²-1)/(x+1)+1/(x+1)��һ����ƽ����лл��Ľ�𣬶��Ұ���ܴ�~~ʮ�ָ�л������