不定积分∫y/1+y²dy怎么求？

Answer 1

∫1/(y2-1) dy =∫1/[(y+1)(y-1)] dy =1/2 ∫[1/(y-1)-1/(y+1)] dy =1/2[∫1/(y-1) dy-∫1/(y+1)dy] =1/2[ln|y-1|-ln|y+1|]+C =1/2 ln|(y-1)/(y+1)|+C

Answer 2

1/2ln｜y²＋1｜