Tricky partial fractions Im trying to do partial fraction but cant seem to get it right, and the examiner have not showed how he did the partial fraction, just the answer.
I want to partial fraction;
$$\frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}$$
My solution so far:
$\frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})} = \frac{A}{s+\frac{1}{2}}+\frac{Bs+D}{(s+\frac{1}{2})^2+\frac{3}{4}} = \frac{A((s+\frac{1}{2})^2+\frac{3}{4})+(Bs+D)(s+\frac{1}{2})}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}$
$\Rightarrow As^2+As+A+Bs^2+\frac{1}{2}Bs+Ds+\frac{1}{2}D=1$
Comparing coefficients gives me
$s^2:A+B=0$
$s^1:A+\frac{1}{2}B+\frac{1}{2}D=0$
$s^0:A+\frac{1}{2}D=1$
Which gives me $A=2, B=-2, C=-2$ which is wrong.
Can someone tell me what I'm doing wrong? Something tells me that its the $Bs+D$ that is wrong.
Thanks!

There are two approaches here - complete the square later, or as Steven Stadnicki suggests, do a change of variable to make your life easier. I'll show my intended solution of completing squares later.
Let $$F(s) = \frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}.$$
We can show that after re-expanding and clearing fractions, we can write $$F(s) = \frac{2}{(2s+1)(s^2 + s + 1)}$$ which makes our lives easier. Now do the usual expansion:
$$\frac{2}{(2s+1)(s^2 + s + 1)} = \frac{A}{2s+1} + \frac{Bs + C}{s^2 + s + 1}$$
and clearing fractions gives
$$2 = A(s^2 + s + 1) + (Bs + C)(2s+1).$$
One can equate coefficients here but I will opt to use the Heaviside cover-up method. With $s = -\frac{1}{2}$ we get $A = \frac{8}{3}.$ For $s = 0$ we obtain the equation $2 = \frac{8}{3} + C$ which gives $C = -\frac{2}{3}.$ Finally, as we have run out of "nice" $s$ values, we'll choose $s = 1$ which will give $B = -\frac{4}{3}$ (check this!). Thus our expansion is
$$F(s) = \frac{\frac{8}{3}}{2s+1} + \frac{-\frac{4}{3}s - \frac{2}{3}}{s^2 + s + 1}$$
and, after re-completing the square and normalizing,
$$F(s) = \frac{\frac{4}{3}}{s+\frac{1}{2}} + \frac{-\frac{4}{3}s - \frac{2}{3}}{\left(s + \frac{1}{2} \right)^2 + \frac{3}{4}}.$$
The inverse Laplace transform should then be easy to compute (an exponential, a sine, and a cosine).
