Evaluate $\int \frac{dt}{(t^2-1)^2}$ I have this problem of evaluating the indefinite integral of a rational function
$$\int \frac{dt}{(t^2-1)^2}$$
and I'm a bit unsure about how to proceed. I could use partial fractions:
$$\frac{1}{(t+1)(t-1)(t+1)(t-1)}$$
$$\frac{1}{(t+1)^2(t-1)^2}$$
$$\frac{A}{t+1} + \frac{B}{t-1} + \frac{C}{t+1} + \frac{D}{(t+1)^2}$$
and go from there... but that seems long and complicated. Is there a trick I'm missing?

$$\frac{1}{t^2-1} = \frac{1}{2}\left(\frac{1}{t-1} - \frac{1}{t+1}\right)\tag{1}$$
Square both sides:
$$\frac{1}{\left(t^2-1\right)^2} = \frac{1}{4}\left(\frac{1}{(t-1)^2} + \frac{1}{(t+1)^2}\right) - \frac{1}{2} \frac{1}{t^2-1}$$
The last term's partial fraction expansion is given by (1), therefore we have:
$$\frac{1}{\left(t^2-1\right)^2} = \frac{1}{4}\left(\frac{1}{(t-1)^2} - \frac{1}{t-1} +\frac{1}{(t+1)^2}+\frac{1}{t+1}\right)$$
