Question

高数凑微分∫1/√(x^2-1)dx=

Answer 1

由题可知x^2>1 x>1或x＜-1
令x=secα，α∈[0，π/2]∪[-π，-π/2]
tanα>0
∫dx/√(x^2-1)=∫dsecα/|tanα|
=∫dsecα/tanα=∫secαtanαdα/tanα=∫secαdα
=ln|secα+tanα|+C=ln|x+√(x^2-1)|+C
所以∫dx/√(x^2-1)=ln|x+√(x^2-1)|+C

追答

令x=secα，α∈(0，π/2)∪(-π，-π/2)

Answer 2

∫dx/√(x^2-1)
let
x=secu
dx=secu.tanu du
∫dx/√(x^2-1)
=∫secu.tanu du/ tanu
=∫secu du
=ln|secu + tanu| + C
=ln|x + √(x^2-1) | + C