Answer
$$2 \ln |x+3|-\ln |x+5|-\frac{2}{3} \ln |3 x-2|+C$$
Work Step by Step
Given $$\int \frac{\left(x^{2}+3 x-44\right) d x}{(x+3)(x+5)(3 x-2)}$$
Since
\begin{align*}
\frac{\left(x^{2}+3 x-44\right)}{(x+3)(x+5)(3 x-2)}&=\frac{A}{(x+3)}+\frac{B}{(x+5)}+\frac{C}{(3 x-2)}\\
&=\frac{A(x+5)(3 x-2)+B(x+3)(3 x-2)+C(x+3)(x+5)}{(x+3)(x+5)(3 x-2)}\\
x^{2}+3 x-44&=A(x+5)(3 x-2)+B(x+3)(3 x-2)+C(x+3)(x+5)
\end{align*}
Then
\begin{align*}
\text{at } x&=-3\ \ \ \ \ A=2 \\
\text{at } x&=-5\ \ \ \ \ B=-1\\
\text{at } x&=2/3\ \ \ \ \ B=-2
\end{align*}
Hence
\begin{aligned}
\int \frac{\left(x^{2}+3 x-44\right) d x}{(x+3)(x+5)(3 x-2)} &=\int \frac{2 d x}{(x+3)}-\int \frac{d x}{(x+5)}-\int \frac{2 d x}{(3 x-2)} \\
&=2 \ln |x+3|-\ln |x+5|-\frac{2}{3} \ln |3 x-2|+C
\end{aligned}
