Question

1/(1-x^2)的不定积分？

哪里错了 三角换元算不对

Answer 1

∫dx/(1-x^2)
=∫dx/[(1-x)(1+x) ]
=(1/2)∫[1/(1-x) +1/(1+x) ] dx
=(1/2) ln| (1+x)/(1-x) | + C

追问

用三角代换呢

追答

x=sinu
dx=cosu du
∫dx/(1-x^2)
=∫[cosu/(cosu)^2] du
=∫ secu du
=ln|secu + tanu| + C
=ln| 1/√(1-x^2) + x/√(1-x^2)| + C
=ln| (1+x)/√(1-x^2) | + C
=ln| √(1+x)/√(1-x) | + C
=(1/2)| (1+x)/(1-x) | + C
那是一样

Answer 2

= ∫ d(x^2) / 2 / (1+ x^2)^1/2
= (1+ x^2)^1/2 + C

Answer 3

解如下图所示

追答