Question

请问怎么求不定积分！如何把分母分子拆开？

想要一份详细过程谢谢

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