Ｉ=∫dx/{(1-x^2)√(1+x^2)}
=∫{(cos(θ)/(2(cos(θ))^2-1}dθ
=∫{(cos(θ)/(2-2(sin(θ))^2-1}dθ
=∫{(cos(θ)/(1-2(sin(θ))^2}dθ

ここでsin(θ)=tとおくと
dt=cos(θ)dθより
{(cos(θ))/(1-2(sin(θ))^2}dθ
={(cos(θ))/(1-2(sin(θ))^2}(dt/cos(θ))
=dt/(1-2t^2)
Ｉ=∫dt/(1-2t^2)
=(1/2)∫dt/(1+(√2)t)+(1/2)∫dt/(1-(√2)t)+C
=(1/(2√2))log|1+(√2)t|-(1/(2√2))log|1-(√2)t|+C
=(1/(2√2))log|(1+(√2)t)/(1-(√2)t)|+C
=(1/(2√2))log|(1+(√2)(x/√(1+x^2)))/(1-(√2)(x/√(1+x^2)))|+C
=((√2)/4)log|(√(1+x^2)+(√2)x)/(√(1+x^2)-(√2)x)|+C