Answer
$$2 \ln |x-2|-\frac{1}{2} \ln |2 x+1|+2 \ln |x+1|+C$$
Work Step by Step
Given $$\int \frac{\left(7 x^{2}+x\right) d x}{(x-2)(2 x+1)(x+1)}$$
Since
\begin{aligned}
\frac{\left(7 x^{2}+x\right)}{(x-2)(2 x+1)(x+1)} &=\frac{A}{(x-2)}+\frac{B}{(2 x+1)}+\frac{C}{(x+1)} \\
&=\frac{A(2 x+1)(x+1)+B(x-2)(x+1)+C(x-2)(2 x+1)}{(x-2)(2 x+1)(x+1)}\\
7 x^{2}+x &=A(2 x+1)(x+1)+B(x-2)(x+1)+C(x-2)(2 x+1)\\
\end{aligned}
Then
\begin{align*}
\text{at } x&=2 \ \ \ \ \to A=2 \\
\text{at } x&=-1/2\ \ \ \ \to B=-1 \\
\text{at } x&=-1\ \ \ \ \to C=2
\end{align*}
Hence
\begin{aligned}
\int \frac{\left(7 x^{2}+x\right)}{(x-2)(2 x+1)(x+1)} d x &=\int \frac{2}{(x-2)} d x-\int \frac{1}{(2 x+1)} d x+\int \frac{2}{(x+1)} d x \\
&=2 \ln |x-2|-\frac{1}{2} \ln |2 x+1|+2 \ln |x+1|+C
\end{aligned}
