求(2x/(1-x^2))的不定积分

Question

求(2x/(1-x^2))的不定积分

Answer 1

如图

Answer 2

∫ (1+2x²)/[x²(1+x²)] dx
=∫ [(1/x²)+1/(1+x²)]dx
=∫ (1/x²) dx+∫ 1/(1+x²) dx
=-1/x+arctanx+C
C为任意常数

Answer 3

∫(1+2x^2)/[x^2*(1+x^2)]dx
=∫(1+x^2+x²)/[x^2*(1+x^2)]dx
=∫(1/(1+x^2)+1/[x^2]dx
=arctanx-1/x+c

Answer 4

∫2x/(1-x^2)dx
=∫(1/(1-x)-1/(1+x))dx
=-ln(x-1)-ln(x+1)+C
=-ln(x²-1)+C

Answer 5

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