Answer
$$\ln |2 x+3|+\ln |x-3|+\frac{3}{(x-3)}+C$$
Work Step by Step
Given $$\int \frac{\left(4 x^{2}-21 x\right) d x}{(x-3)^{2}(2 x+3)}$$
Since
\begin{align*}
\frac{ 4 x^{2}-21 x }{(x-3)^{2}(2 x+3)}&=\frac{A}{(2 x+3)}+\frac{B}{(x-3)}+\frac{C}{(x-3)^{2}}\\
&= \frac{ A(x-3)^{2}+B \cdot(2 x+3)(x-3)+C(2 x+3)}{(x-3)^{2}(2 x+3)}\\
4 x^{2}-21 x&= A(x-3)^{2}+B \cdot(2 x+3)(x-3)+C(2 x+3)
\end{align*}
Then
\begin{align*}
\text{at } x&=-3/2 \ \ \ \ \to A=2 \\
\text{at } x&= 3\ \ \ \ \to C=-3\\
\text{at } x&= 0\ \ \ \ \to B=1
\end{align*}
Hence
\begin{aligned}
\int \frac{\left(4 x^{2}-21 x\right) d x}{(x-3)^{2}(2 x+3)} &=\int \frac{2 d x}{(2 x+3)}+\int \frac{d x}{(x-3)}-\int \frac{3 d x}{(x-3)^{2}}\\
&=\ln |2 x+3|+\ln |x-3|+\frac{3}{(x-3)}+C
\end{aligned}
