将 $\dfrac{-6x^3 - 9x^2 + x - 16}{(x^2 - 1)(2x^2 + 3)}$ 化为部分分式。

1. Factorise the denominator and set up the correct partial fraction form.

Since
$x^2-1=(x-1)(x+1),$
write
$\frac{-6x^3-9x^2+x-16}{(x^2-1)(2x^2+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{2x^2+3}.$

2. Clear the denominator to form a polynomial identity.

Multiplying throughout by $(x-1)(x+1)(2x^2+3)$,
$-6x^3-9x^2+x-16=A(x+1)(2x^2+3)+B(x-1)(2x^2+3)+(Cx+D)(x^2-1).$

3. Expand the right-hand side and collect like powers of $x$.

Expand:
$A(x+1)(2x^2+3)=A(2x^3+2x^2+3x+3),$
$B(x-1)(2x^2+3)=B(2x^3-2x^2+3x-3),$
$(Cx+D)(x^2-1)=Cx^3+Dx^2-Cx-D.$
Hence the right-hand side is
$(2A+2B+C)x^3+(2A-2B+D)x^2+(3A+3B-C)x+(3A-3B-D).$

4. Compare coefficients to obtain simultaneous equations.

Comparing coefficients of powers of $x$:
$2A+2B+C=-6,$
$2A-2B+D=-9,$
$3A+3B-C=1,$
$3A-3B-D=-16.$

5. Solve for the constants $A$, $B$, $C$ and $D$.

Adding the first and third equations:
$(2A+2B+C)+(3A+3B-C)=-6+1,$
$5A+5B=-5,$
$A+B=-1.$
From $3A+3B-C=1$,
$3(-1)-C=1,$
$C=-4.$
Let $A+B=-1$. Using the second and fourth equations:
$2A-2B+D=-9,$
$3A-3B-D=-16.$
Adding them,
$5A-5B=-25,$
$A-B=-5.$
Solving
$A+B=-1,\qquad A-B=-5,$
gives
$A=-3,\qquad B=2.$
Then from $2A-2B+D=-9$,
$2(-3)-2(2)+D=-9,$
$D=1.$

6. Substitute the constants into the partial fraction form.

Therefore,
$A=-3,\quad B=2,\quad C=-4,\quad D=1.$
So
$\frac{-6x^3-9x^2+x-16}{(x^2-1)(2x^2+3)}=-\frac{3}{x-1}+\frac{2}{x+1}+\frac{-4x+1}{2x^2+3}.$