If $f(x)=a \log |x|+b x^2+x$ has its extremum values at $x=-1$ and $x=2$, then

$a=2, b=-1 / 2$

We have,

$f(x)=a \log |x|+b x^2+x \Rightarrow f^{\prime}(x)=\frac{a}{x}+2 b x+1$

Since f(x) attains its extremum values at x = -1, 2

∴  $f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$

$\Rightarrow -a-2 b+1=0$ and $\frac{a}{2}+4 b+1=0$

$\Rightarrow a=2$ and $b=-1 / 2$