求不定积分1⼀x √(1+x^2)dx

2025-05-23 22:14:59
推荐回答(1个)
回答1:

设x=tanu,du=dx/(1+x^2)
∫1/x √(1+x^2)dx
=∫cotudu
=ln|sinu|+c
=ln|x/√(1+x^2)|+c