1/x（根号下（a^2-x^2))求不定积分

Question

1/x（根号下（a^2-x^2))
求不定积分

Answer 1

∫1/[x√(a^2-x^2)]dx
= (1/a^2)∫ [ √(a^2-x^2)/ x + x/√(a^2-x^2) ] dx
=(1/a^2)[ ∫ √(a^2-x^2)/ x dx - ∫ d√(a^2-x^2) ]
= (1/a^2) ∫ √(a^2-x^2)/ x dx - √(a^2-x^2)/(a^2)
let
a/x = secb
(-a/x^2) dx = (tanb)^2db
(-a/(a/secb)^2) dx = (tanb)^2db
dx = -a (sinb)^2 db
∫ √(a^2-x^2)/ x dx
= ∫ tanb[ -a (sinb)^2 db ]
= -a∫ (sinb)^3/cosb db
= a ∫ (1-(cosb)^2)/cosb dcosb
= a [ln|cosb| - (cosb)^2/2 ] + C'
= a[ln|x/a| - (1/2)(x/a)^2] + C'
∫1/[x√(a^2-x^2)]dx
= (1/a^2) ∫ √(a^2-x^2)/ x dx - √(a^2-x^2)/(2a^2)
= (1/a)[ln|x/a| - (1/2)(x/a)^2 ] -√(a^2-x^2)/(a^2) + C