well

write as dx/x(1-x^2) =dx/x(1-x)(1+x)

[1/3{ 2/x(1+x) + 1/(1-x)(1+x) + 1/x(1-x)}]dx

= {1/x -1/2*1/1+x + 1/2*1/1-x}dx

intergrating v get:

lnx-1/2ln(1+x) -1/2ln(1-x) +c

= 1/2{ 2lnx  - ln 1-x^2} +c

=1/2{ln(x^2/1-x^2)} +c