求x/（x²-x-2）的不定积分

Question

如题所述

Answer 1

简单计算一下即可，答案如图所示

Answer 2

方法如下，
请作参考：

Answer 3

x²-x-2=(x+1)(x-2)，设x/(x²-x-2)=a/(x-2)+b/(x+1)。解得a=2/3，b=1/3。

∴原式=(1/3)∫[2/(x-2)+1/(x+1)]dx=(2/3)ln丨x-2丨+(1/3)ln丨x+1丨+C。

Answer 4

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