Answer
$$\frac{1}{4} \ln \left|\frac{3}{4}\right|+\frac{37}{480}$$
Work Step by Step
Given $$\int_{4}^{9} \frac{d t}{\left(t^{2}-1\right)^{2}}$$
Let $$t=\sec u\ \ \ \ \ \ \ dt=\sec u\tan udu $$
\begin{align*}
\text{at }\ t&= 4\ \ \ \ \ \ \ u =\sec^{-1}(4)\\
\text{at }\ t&= 9\ \ \ \ \ \ \ u =\sec^{-1}(9)
\end{align*}
Then
\begin{align*}
\int_{4}^{9} \frac{d t}{\left(t^{2}-1\right)^{2}}&=\int_{\sec^{-1}(4)}^{\sec^{-1}(9)} \frac{\sec u\tan udu}{\left(\sec^{2}u-1\right)^{2}}\\
&=\int_{\sec^{-1}(4)}^{\sec^{-1}(9)} \frac{\sec u\tan udu}{\tan^{4}u}\\
&=\int_{\sec^{-1}(4)}^{\sec^{-1}(9)} \frac{\sec u du}{\tan^{3}u}\\
&=\int_{\sec^{-1}(4)}^{\sec^{-1}(9)}\csc u\cot^2udu\\
&=\int_{\sec^{-1}(4)}^{\sec^{-1}(9)}(\csc^3 u-\csc u)du\\
&=\frac{1}{4}\left(\frac{\tan ^2\left(\frac{u}{2}\right)}{2}-2\ln \left|\tan \left(\frac{u}{2}\right)\right|-\frac{1}{2\tan ^2\left(\frac{u}{2}\right)}\right)\bigg|_{\sec^{-1}(4)}^{\sec^{-1}(9)}\\
&=\frac{1}{4} \ln \left|\frac{3}{4}\right|+\frac{37}{480}
\end{align*}
