Answer
$\dfrac{1}{4}\ln \left|t+1\right|-\dfrac{1}{4\left(t+1\right)}-\dfrac{1}{4}\ln \left|t-1\right|-\dfrac{1}{4\left(t-1\right)}+C$
Work Step by Step
$\displaystyle \frac{1}{(t^{2}-1)^{2}}=\frac{1}{[(t+1)(t-1)]^{2}}=$
$=\displaystyle \frac{1}{(t+1)^{2}(t-1)^{2}}=\frac{A}{t+1}+\frac{B}{(t+1)^{2}}+\frac{C}{t-1}+\frac{D}{(t-1)^{2}}$
$1=A(t+1)(t -1)^{2}+B(t -1)^{2}+C(t+1)^{2}(t -1)+D(t+1)^{2}$
$t=1\quad \Rightarrow\quad 4D=1\Rightarrow\quad D=1/4$
$t=-1\quad \Rightarrow\quad 4B=1\Rightarrow\quad B=1/4$
$t=0\quad \Rightarrow\quad\left\{\begin{array}{ll}
1= A+B-C+D & \\
1=A+\frac{1}{4}-C+\frac{1}{4} & \\
& \\
A-C=\frac{1}{2} & (*)
\end{array}\right.$
$t=2\quad \Rightarrow\quad\left\{\begin{array}{ll}
1= 3A+B+9C+9D & \\
1=3A+\frac{1}{4}+9C+\frac{9}{4} & \\
3A+9C=\frac{-3}{2} & \\
A+3C=-\frac{1}{2} & (**)
\end{array}\right.$
Subtract (**) from (*)
$-4C=1$
$C=-\displaystyle \frac{1}{4}$
substsitute into (*) $\Rightarrow A=\displaystyle \frac{1}{4}.$
$\displaystyle \int\frac{1}{(t^{2}-1)^{2}}dt=\int(\frac{\frac{1}{4}}{t+1}+\frac{\frac{1}{4}}{(t+1)^{2}}+\frac{-\frac{1}{4}}{t-1}+\frac{\frac{1}{4}}{(t-1)^{2}})dt$
$=\dfrac{1}{4}\ln \left|t+1\right|-\dfrac{1}{4\left(t+1\right)}-\dfrac{1}{4}\ln \left|t-1\right|-\dfrac{1}{4\left(t-1\right)}+C$
