設(shè)1/(x^2-1)^2=[(ax+b)/(x-1)^2+(cx+d)/(x+1)^2],
右邊通分,對(duì)應(yīng)項(xiàng)相等,得到：
a=-1/4,c=1/4,b=d=1/2.
所以積分為：
原式子
=-（1/4)∫(x-2)dx/(x-1)^2+(1/4)∫(x+2)dx/(x+1)^2
=(1/4)∫xd[1/(x-1)]+(1/2)∫dx/(x-1)^2-(1/4)∫xd[(1/(x+1)]+(1/2)∫dx/(x+1)^2
=(1/4)x/(x-1)-(1/4)∫dx/(x-1)-[1/(2x-2)]-(1/4)x/(x+1)+(1/4)∫dx/(x+1)-[1/(2x+2)]
=(1/4)x/(x-1)-(1/4)x/(x+1)-(1/4)ln|(x-1)/(x+1)|-[1/(2x-2)+1/(2x+2)]+c.后面還可以做通分適當(dāng)化簡(jiǎn).