Answer
$\frac{1}{4}{\ln|t+1|-\frac{1}{4(t+1)}-\frac{1}{4}{\ln|t-1|}-\frac{1}{4(t-1)}}+C$
Work Step by Step
$\int{\frac{dt}{(t^2-1)^2}}$ = $\int{\frac{dt}{(t-1)^2(t+1)^2}}$ = $\int\left[\frac{A}{t+1}+\frac{B}{(t+1)^2}+\frac{C}{t-1}+\frac{D}{(t-1)^2}\right]dt$
$1$ = $A(t+1)(t-1)^2+B(t-1)^2+C(t-1)(t+1)^2+D(t+1)^2$
$A$ = $\frac{1}{4}$, $B$ = $\frac{1}{4}$, $C$ = $-\frac{1}{4}$, $D$ = $\frac{1}{4}$
So
$\int{\frac{dt}{(t^2-1)^2}}$ = $\int\left[\frac{1}{4(t+1)}+\frac{1}{4(t+1)^2}-\frac{1}{4(t-1)}+\frac{1}{4(t-1)^2}\right]dt$ = $\frac{1}{4}{\ln|t+1|-\frac{1}{4(t+1)}-\frac{1}{4}{\ln|t-1|}-\frac{1}{4(t-1)}}+C$
