1⼀（x^2-2x-3）的不定积分怎么求?

2024-11-08 11:03:16

推荐回答（2个）

回答1：

∫[1/(x²-2x-3)]dx
=∫[1/(x+1)(x-3)]dx
=¼∫[(x+1)-(x-3)]/[(x+1)(x-3)] dx
=¼∫[1/(x-3) -1/(x+1)]dx
=¼∫[1/(x-3)]d(x-3) -¼∫[1/(x+1)]d(x+1)
=¼ln|x-3|-¼|ln(x+1)|+C
=¼ln|(x-3)/(x+1)| +C

回答2：

1/(x^2-2x-3) = (1/4)[1/(x-3) -1/(x+1)]

∫dx/(x^2-2x-3)
=(1/4)∫[1/(x-3) -1/(x+1)] dx
=(1/4) ln|(x-3)/(x+1)| + C