Question

求不定积分∫1/x√(x^2-9)dx

Answer 1

设x=3sect,dx=3sect*tantdt,
cost=3/x,t=arccos(3/|x|),
tant=√[(sect)^2-1]=√(x^2/9-1)
原式=∫sect*tantdt/(|tant|*3sect)
=(1/3)∫dt
=t/3+C
=(1/3)arccos(3/|x|)+C

Answer 2

x=3sect dx=3secttantdt √(x^2-9)=3tant
∫1/x√(x^2-9)dx
=∫{1/[(3sect)3tant]}3secttantdt
=(1/3)∫dt
=(1/3)t +c
=(1/3)arccos3/x +c