求不定积分∫1/(x²-x-2)dx_数学解答

求不定积分∫1/(x²-x-2)dx

人气：279 ℃　时间：2019-12-13 15:53:16

解答

∫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