Answer
$$-\frac{1}{49} \ln |t-3|-\frac{1}{7(t-3)}+\frac{1}{49} \ln |t+4|+C$$
Work Step by Step
Given $$ \int \frac{d t}{(t-3)^{2}(t+4)}$$
Since
\begin{aligned}
\frac{1}{(t-3)^{2}(t+4)} &=\frac{A}{(t-3)}+\frac{B}{(t-3)^{2}}+\frac{C}{(t+4)} \\
&=\frac{A(t-3)(t+4)+B(t+4)+C(t-3)^{2}}{(t-3)^{2}(t+4)} \\
1 &=A(t-3)(t+4)+B(t+4)+C(t-3)^{2}
\end{aligned}
Then
\begin{align*}
\text{at } x&=3 \ \ \ \ \to A=\frac{1}{7} \\
\text{at } x&= -4\ \ \ \ \to C= \frac{1}{49}\\
\text{at } x&= 04\ \ \ \ \to B= \frac{1}{49}
\end{align*}
Then
\begin{aligned}
\int \frac{d t}{(t-3)^2(t+4)} &=-\frac{1}{49} \int \frac{1}{(t-3)} d t+\frac{1}{7} \int \frac{1}{(t-3)^{2}} d t+\frac{1}{49} \int \frac{1}{(t+4)} d t \\
&=-\frac{1}{49} \int \frac{1}{(t-3)} d t+\frac{1}{7} \int(t-3)^{-2} d t+\frac{1}{49} \int \frac{1}{(t+4)} d t \\
&=-\frac{1}{49} \ln |t-3|+\frac{1}{7} \cdot \frac{(t-3)^{-1}}{(-1)}+\frac{1}{49} \ln |t+4|+C \\
&=-\frac{1}{49} \ln |t-3|-\frac{1}{7(t-3)}+\frac{1}{49} \ln |t+4|+C
\end{aligned}
