不定积分dx⼀x√(x^2-1)怎么求

arccotx1⼀x+C
2025-06-24 18:52:59
推荐回答(2个)
回答1:

解题关键:第二类换元积分法。

满意请采纳!!!

回答2:

∫dx/x√(x^2-1)=1/2∫dx^2/x^2√(x^2-1)= 1/2∫dt/t√(t-1) (令x^2=t)
=1/2∫2udu/(u^2+1)u (令√(t-1) =u,t=u^2+1)
=∫du/(u^2+1)
=actanu+C
=actan√(x^2-1)+C