求∫(x^2+1)/[(x+1)^2(x-1)]dx

Question

求∫(x^2+1)/[(x+1)^2(x-1)]dx

Answer 1

原式=（1/2）∫[（x-1）/(x+1)^2+1/(x-1)]dx
=(1/2)∫[(x+1-2)/(x+1)^2+1/(x-1)]dx
=(1/2)∫[1/(x+1)-2/(x+1)^2+1/(x-1)]dx
=(1/2)[ln(x+1)+2/(x+1)+ln(x-1)]+C
=(1/2)[ln(x+1)+ln(x-1)]+1/(x+1)+C
之所以拆成第一个等号,是因为设原式=∫[（ax+b）/(x+1)^2+c/(x-1)]dx,对于中括号里的部分通分,从而分子变为（a+c）x^2+(b-a+2c)x+c-b,而原来的分子是x^2+1,所以a+c=1,b-a+2c=0,c-b=1,解得a=c=1/2,b=-1/2