I want to integrate the density of the log logistic distribution (https://en.wikipedia.org/wiki/Log-logistic_distribution). The Wikipedia article already tells us (from the expected value) that -
\begin{equation}\int_0^\infty x.\frac{\frac{\beta}{\alpha} \left(\frac{x}{\alpha}\right)^{\beta - 1}}{(1 + \left(\frac{x}{\alpha}\right)^\beta)^2} = \frac{\alpha \pi}{\beta sin\left(\frac{\pi}{\beta}\right)}\end{equation}
What I want to find is -
$$\int_0^t x.\frac{\frac{\beta}{\alpha} \left(\frac{x}{\alpha}\right)^{\beta - 1}}{(1 + \left(\frac{x}{\alpha}\right)^\beta)^2}dx$$
I tried to use integration by parts, observing that -
$$\frac{-\beta. x^{\beta-1}}{\alpha^\beta(1+\left(\frac{x}{\alpha}\right)^\beta)} = d\left(\frac{1}{(1 + \left(\frac{x}{\alpha}\right)^\beta)^2}\right)$$
So,
$$\int_0^t x.\frac{\frac{\beta}{\alpha} \left(\frac{x}{\alpha}\right)^{\beta - 1}}{(1 + \left(\frac{x}{\alpha}\right)^\beta)^2}dx = -\int_0^t d\left(\frac{1}{(1 + \left(\frac{x}{\alpha}\right)^\beta)^2}\right).x$$
$$ = -\left[\frac{x}{(1 + \left(\frac{x}{\alpha}\right)^\beta)}\right]_0^t + \int_0^t \frac{dx}{(1 + \left(\frac{x}{\alpha}\right)^\beta)}$$
Now, I'm stuck on the second part of the above integral. We know that as $t \to \infty$ it becomes the expected value (since the first part then becomes $0$). But is there a closed form for the bounded integral as well? If not, is there a good numerical way to solve it?