∫1/(1-x^2)dx
=∫1/[(1+x)(1-x)]dx
=1/2·∫[1/(1+x)+1/(1-x)]dx
=1/2·[ln|1+x|+ln1-x|]+C
=1/2·ln|(1+x)(1-x)|+C
令x=tanu,則dx=(secu^2) du
∫1/√(1+x^2)dx
=∫1/secu·(secu)^2 du
=∫secu du
=ln|tanu+secu|+C
=ln|x+√(1+x^2)|+C