#### Answer

$\displaystyle \dfrac{1}{(x-2)(x^2+2x+4)}=\dfrac{\frac{1}{12}}{x-2}+\dfrac{-\frac{1}{12}x-\frac{1}{3}}{x^2+2x+4} $

#### Work Step by Step

We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has only non-repeated irreducible quadratic factor.
So, we will write the given rational expression into partial fraction form as:
$$\displaystyle \frac{1}{(x-2)(x^2+2x+4)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4} \quad \quad(1)$$
Now, write the RHS with a common denominator.
\begin{align*}\displaystyle \frac{1}{(x-2)(x^2+2x+4)}&=\frac{A(x^2+2x+4)+(x-2)(Bx+C)}{(x-2)(x^2+2x+4)}\\
\\\frac{1}{(x-2)(x^2+2x+4)}&=\frac{Ax^2+2Ax+4A+Bx^2-2Bx+Cx-2C}{(x-2)(x^2+2x+4)}
\\\frac{1}{(x-2)(x^2+2x+4)}&=\frac{(A+B)x^2+(2A-2B+C)x+(4A-2C)}{(x-2)(x^2+2x+4)}
\end{align*}
Next, equate the numerators to obtain:
$$(A+B)x^2+(2A-2B+C)x+(4A-2C)=1$$
Equate the coefficients of the polynomials on the LHS and RHS to obtain:
$$A+B=0 \quad \quad\ (2)
\\ 2A-2B+C=0 \quad \quad (3)
\\4A-2C=1\quad \quad(4)$$
Equation $(2)$ yields: $A=-B$.
Plug $A=-B$ into the Equation $(3)$.
$$2(-B)-2B+C=1\\ -2B-2B+C=0\\ -4B+C=0 \\ C=4B ~~~~(5)$$
Next, plug $A=-B$ and $C=4B$ into the Equation $(4)$.
$$4(-B)-2(4B)=1\\ -4B-8B=1\\ -12B=1 \\ B=\dfrac{-1}{12}$$
Hence, $A=-B=-(\dfrac{-1}{12})=\dfrac{1}{12}$
Finally, plug $B=-\frac{1}{12}$ into Equation $(5)$ to solve for $C$.
$$C =4(\dfrac{-1}{12})=\dfrac{-1}{3}$$
Therefore, equation $(1)$ becomes:
$$\displaystyle \dfrac{1}{(x-2)(x^2+2x+4)}=\dfrac{\frac{1}{12}}{x-2}+\dfrac{\frac{-1}{12}x+\dfrac{-1}{3}}{x^2+2x+4} $$
or, $$\displaystyle \dfrac{1}{(x-2)(x^2+2x+4)}=\dfrac{\frac{1}{12}}{x-2}+\dfrac{-\frac{1}{12}x-\frac{1}{3}}{x^2+2x+4} $$