Evaluate: `int1/((1+x^2)sqrt(1-x^2))dx`

Explanation

Putting `x=(1)/(t)` and `dx=-(1)/(t^(2))dt,` we get
`I=int((-(1)/(t^(2)))dt)/((1-(1)/(t^(2)))sqrt(1+(1)/(t^(2))))=-int(tdt)/((t^(2)-1)sqrt(t^(2)+1))`
Let `t^(2)+1=u^(2),` or `2tdt=2udu`
`:. I=-int(du)/(u^(2)-(sqrt(2))^(2))`
`=-(1)/(2sqrt(2))log|(u-sqrt(2))/(u+sqrt(2))|+C`
`=-(1)/(2sqrt(2))log|(sqrt(t^(2)+1)-sqrt(2))/(sqrt(t^(2)+1)+sqrt(2))|+C`
`=-(1)/(2sqrt(2))log|(sqrt((1)/(x^(2))+1)-sqrt(2))/(sqrt((1)/(x^(2))+1)+sqrt(2))|+C`
`=-(1)/(2sqrt(2))log|(sqrt(1+x^(2))-sqrt(2)x)/(sqrt(1+x^(2))+sqrt(2)x)|+C`