Question

高数，谢谢啦

Answer 1

Answer 2

let
t= tanu
dt = (secu)^2 du

∫√(1+t^2)/ t^2 dt
=∫ [(secu)^3/(tanu)^2 ] du
=∫ du/[ (sinu)^2.cosu]
=∫ [1/cosu + cosu/(sinu)^2] du
=∫ secu du + ∫[cosu/(sinu)^2] du
=ln|secu+tanu| - 1/sinu + C
=ln|√(1+t^2)+t | - √(1+t^2)/t + C

Answer 3

令t=tanu
则dt=sec²udu
原式=∫sec³u/tan²u·du
=∫secu(1+tan²u)/tan²u·du
=∫secudu+∫secu/tan²u·du
=ln|secu+tanu|+∫cscucotudu
=ln|secu+tanu|-cscu+C
=ln|x+√(x²+1)|-√(x²+1)/x+C

追答

无语，错的答案也可以采纳

sb，祝你期末考试挂科快乐