题目

高数不定积分问题!
求不定积分：∫sec³xdx.
∫dx/x²（1-x^4).

提问时间：2021-02-01

答案

(1)∫sec³xdx
=∫sec²xsecxdx
=∫(1+tan²x)secxdx
=∫secxdx+∫tan²xsecxdx
=∫secxdx+∫tanxd(secx)
=∫secxdx+secxtanx-∫secxd(tanx)
=∫secxdx+secxtanx-∫sec³xdx
∵2∫sec³xdx=∫secxdx+secxtanx
∴∫sec³xdx=(1/2)(ln|secx+tanx|+secxtanx)+C
(2)∫dx/[x²(1-x^4)]
=∫[1/x²+(1/4)/(x+1)-(1/4)/(x-1)-(1/2)/(x²+1)]
=∫dx/x²+(1/4)∫dx/(x+1)-(1/4)∫dx/(x-1)-(1/2)∫dx/(x²+1)
=-1/x+(1/4)/ln(x+1)-(1/4)/ln(x-1)-(1/2)arctanx+C