求不定积分∫1/(x²-x-2)dx

求不定积分∫1/(x²-x-2)dx

∫1/(x²-x-2)dx
=∫1/[(x-2)(x+1)]dx
=1/3∫[1/(x-2)-1/(x+1)]dx
=1/3ln(x-2)-1/3ln(x+1)+C

由题意可得：
∫1/(x²-x-2)dx
=∫1/[(x-2)(x+1)]dx
=-∫[1/(x-2)+1/(x+1)]dx
=-∫1/(x-2)dx-∫1/(x+1)dx
=-∫1/(x-2)d(x-2)-∫1/(x+1)d(x+1)
=-ln|x-2|-ln|x+1|+C (C为任意常数)

1/(x^2-x-2)=1/(x-2)(x+1)
令它为a/(x-2)+b/(x+1)=(ax+a+bx-2b) /(x-2)(x+1)
a+b=0 a-2b=1 得到 a=1/3 b=-1/3
1/(x^2-x-2)=1/3(x-2)-1/3(x+1)
∫1/(x²-x-2)dx
=∫1/3(x-2)-1/3(x+1)dx
=1/3*ln(x-2)-1/3*ln(x+1)+C