对(1-x^4)^-1积分?

∫1/(1-x^4)dx
=∫1/(1+x²)(1-x²)dx
=[∫1/(1+x²)+1/(1-x²)dx]/2
=[∫1/(1+x²)dx]/2+[∫1/(1-x²)dx]/2
=[∫1/(1+x²)dx]/2+[∫1/(1+x)(1-x)dx]/2
=[∫1/(1+x²)dx]/2+[∫1/(1+x)+1/(1-x)dx]/4
=[∫1/(1+x²)dx]/2+[∫1/(1+x)dx]/4-[∫1/(1-x)d(-x)]/4
=arctanx/2+ln(1+x)/4-ln(1-x)/4+C