∫x눀/(1-x눀)dx?

求大神解答૶૶
2025-06-27 23:28:41
推荐回答(3个)
回答1:

望采纳

回答2:

不定积分计算过程如下:
∫x²/(1-x²)dx
=∫[1-(1-x²)]/(1-x²)dx
=∫1/(1-x²)dx-∫dx
=ln|(1+x)/(x-1)|-x+c

回答3:

∫(√(1-x²))/ x² dx
let
x = sina
dx = cosa da
∫(√(1-x²))/ x² dx
=∫ [cosa/(sina)^2] cosa da
=∫ (cota)^2 da
=∫ [(csca)^2-1] da
=-cota -a + C
= -√(1-x²) /x - arcsinx + C