Question

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

令x=sinz，dx=coszdz
∫dx/[x+√(1-x²)]
=∫cosz/(sinz+cosz)*dz
=(1/2)∫[(sinz+cosz)+(-sinz+cosz)]/(sinz+cosz)*dz
=(1/2)∫dz + (1/2)∫d(cosz+sinz)/(sinz+cosz)*dz
=(1/2)z + (1/2)ln|sinz+cosz| + C
=(1/2)(arcsinz) + (1/2)ln|x+√(1-x²)| + C

Answer 3

左边是lnx，右边令x=sint,则 √1-x^2dt=(cost)^2dt=(cos2t+1)/2dt
所以1/x+√1-x^2的不定积分是lnx+(sin2t)/2+x/2+C

Answer 4

分开嘛
左边是lnx，右边令x=sint,则 √1-x^2dt=(cost)^2dt=(cos2t+1)/2dt
所以1/x+√1-x^2的不定积分是lnx+(sin2t)/2+x/2+C C为常数