$ \frac{5x^3+12x^2+12x+10}{(x^2+4)(x^2+2x+1)}$

$ =\frac{5x^3+12x^2+12x+10}{(x^2+4)(x+1)^2}$

Partial fractions attempt:

$ \frac{Ax+B}{x^2+4}$ + $\frac{C}{x+1}$ + $\frac{D}{(x+1)^2}$

$(Ax+B)(x+1)^2=(Ax+B)(x^2+2x+1)=Ax^3+2Ax^2+Ax+Bx^2+2Bx+B$

$C(x^2+4)(x+1)=C(x^3+x^2+4x+4)=Cx^3+Cx^2+4Cx+4C$

$D(x^2+4)=Dx^2+4D$

$x^3(A+C)=5$

$x^2(2A+B+C+D)=12$

$x^1(A+2B+4C+4D)=12$

$x^0(B+4C+4D)=10$

$A=\frac{32}{5}$

$B=-\frac{22}{5}$

$C=-\frac{7}{5}$

$D=5$

Is this wrong so far? Mathematica seems to think so.