Answer
$$\frac{1}{7}\ln \frac{24}{10}$$
Work Step by Step
Given $$\int_{4}^{6} \frac{d t}{(t-3)(t+4)}$$
Since
\begin{aligned}
\frac{1}{(t-3)(t+4)} &=\frac{A}{(t-3)}+\frac{B}{(t+4)} \\
&=\frac{A(t+4)+B(t-3)}{(t-3)(t+4)} \\
1 &=A(t+4)+B(t-3)
\end{aligned}
\begin{align*}
\text{at }\ t&= -4\ \ \ \ \ \ \ A=\frac{1}{7}\\
\text{at }\ t&= 3\ \ \ \ \ \ \ B=\frac{-1}{7}
\end{align*}
Then
\begin{aligned}
\int_{4}^{6} \frac{d t}{(t-3)(t+4)} &=\frac{1}{7} \int_{4}^{6} \frac{1}{(t-3)} d t-\frac{1}{7} \int_{4}^{6} \frac{1}{(t+4)} d t \\
&=\frac{1}{7} \ln |t-3|-\frac{1}{7} \ln |t+4|\bigg|_{4}^{6}\\
&= \frac{1}{7} \ln |3|-\frac{1}{7} \ln |10|+\frac{1}{7} \ln |8|\\
&=\frac{1}{7}\ln \frac{24}{10}
\end{aligned}
