For 1/((s^2+1)(s^2+2s+3)).

I would first break the original expression into partial fractions.

1/((s^{2}+1)*(s^{2}+2s+3))=(A+Bs)/(s^{2}+1)+(C+Ds)/(s^{2}+2s+3). Now we need to find A, B, C, and D. Since two expressions must be identical, numerators shall be equal on the left and right parts (denominators are equal, since
common denominator on the right is the same as on the left side). Therefore, (A+Bs)(s^{2}+2s+3)+(C+Ds)*(s^{2}+1)=1 or Bs^{3}+(A+2B)s^{2}+(2A+3B)s+3A+Ds^{3}+Cs^{2}+Ds+C=1. Thus we have the following system of
equations:

B+D=0; A+2B+C=0; 2A+3B+D=0; 3A+C=1; B=-D from the first, then we are down to three equations:

A+2B+C=0; 2A+3B-B=0; 3A+C=1. From the second of these three A=-B; Therefore, -B+2B+C=0 and -3B+C=1;

B+C=0 and -3B+C=1; From the first of two B=-C, therefore the second equation gives 4C=1 or C=¼. Therefore, B=-¼; A=¼, and D=¼. So we obtain:

1/((s2+1)*(s2+2s+3))=¼[(s-1)/(s^{2}+1)+(s+1)/(s^{2}+2s+3)]

Now we consider each term in brackets separately.

(s-1)/(s^{2}+1)=s/(s^{2}+1)-1/(s^{2}+1). Using linearity of Laplace transform, we can apply inverse transform to each term separately and obtain: F[s/(s^{2}+1)](t)=cos(t); F[1/(s^{2}+1)](t)=sin(t); therefore, F[(s-1)/(s^{2}+1)](t)=cos(t)-sin(t).

(s+1)/(s^{2}+2s+3)=(s+1)/(s^{2}+2s+1+2)=(s+1)/((s+1)^{2}+2); F[(s+1)/((s+1)^{2}+2)](t)=e^{t}*cos(t√2)

Combining everything together, we obtain:

F[1/((s^{2}+1)*(s^{2}+2s+3))]=¼[cos(t)-sin(t)+e^{t}*cos(t√2)]