求不定积分 ∫√（4+x²）dx,

这个用换元法
令x=2tana
dx=2sec^2ada
∫√（4+x²）dx
=∫√（4+4tan^2a）dx
=∫2seca*2sec^2ada
=4∫secadtana
=4secatana-4∫tanadseca
=4secatana-4∫tan^2a*secada
=4secatana-4∫(sec^2a-1)*secada
所以
4∫sec^3ada=4secatana-4∫sec^3ada+∫secada
8∫sec^3ada=4secatana+∫secada
=4secatana+∫1/cos^2adsina
=4secatana+∫1/(1-sin^2a)dsina
=4secatana+∫[1/(1-sina)+1/(1+sina)]dsina
=4secatana+ln[(1+sina)/(1-sina)+C1
∫√（4+x²）dx=4∫sec^3ada
=1/2secatana+1/4ln[(1+sina)/(1-sina)+C
感觉过程有点别扭