求不定积分的全过程 ∫(1/(x+(1-x²)½))dx

Question

如题所述

Answer 1

先分母有理化，再换元。见图片

Answer 2

设x=sina
原式=∫1/(sina+cosa)dsina
=∫cosa/(sina+cosa)da
=∫cosa(cosa-sina)/(cos^2a-sin^2a)da
=1/2∫(1-cos2a-sin2a)/cos2a da
=1/4∫1/cos2ad2a-a/2+1/4∫1/cos2adcos2a
=1/4∫1/sin2ad2a-a/2+1/4ln|cos2a|
=1/4∫1/2sinacosad2a-a/2+1/4ln|cos2a|
=1/4∫1/tana d(tana)-a/2+1/4ln|cos2a|
=1/4ln|tana|-a/2+1/4ln|cos2a| +c
x=sina
cosa=√(1-x^2)
cos2a=2cos^2a-1=1-x^2-1=-x^2
原式=1/4ln|x/√(1-x^2)|+1/4lnx^2+1/2arcsinx+c
=1/4ln|x/√(1-x^2)|+1/2ln|x|+1/2arcsinx+c