1) (x/t)*(e^(-x/t))*(1+e^(-x/t))^-2

2) ((x^2)/t)*(e^(-x/t))*(1+e^(-x/t))^-2

Both on negative infinity to positive infinity.

You can see that they are both expectations of functions of a weibull random variable with theta of t and beta of 1.

1) Y = X(1+e^(-X/t))^-2

2) Y = (X^2)(1+e^(-X/t))^-2

I'm not sure if you need to use that or if I just can't solve the integrals properly. Any help?