1/(x+√(1-x^2))的不定积分

Answer 1

令a=1即可，原式=

(1/2)arcsinx+(1/2)ln|x+√(1-x²)|+C

Answer 2

设sinu=x,dx=cosudu
原式=∫cosudu/(sinu+ cosu)
=∫dsinu/(sinu+ cosu)
=sinu/(sinu+ cosu)-∫sinud(sinu +cosu)
=sinu/(sinu +cosu)-∫sinu(cosu-sinu)du
=sinu/(sinu+ cosu)-∫sinucosudu +∫sin^2udu
=sinu/(sinu +cosu) +1/4*cos2u+ u/2-1/4*sin2u
=x/(x +(1-x^2)^(1/2))+ (1-x^2)/2-1/4 arcsinu/2-x/2*(1-x^2)^(1/2) +C