Q15Advanced Engineering Mathematics

To evaluate the inverse Laplace transform $L^{-1}\left\{\frac{1}{s(s^2+1)}\right\}$, we first decompose the rational function using partial fractions: $\frac{1}{s(s^2+1)}=\frac{A}{s}+\frac{Bs+C}{s^2+1}$.

Solving for the coefficients yields $A=1$, $B=-1$, and $C=0$. This gives $\frac{1}{s}-\frac{s}{s^2+1}$. Applying the standard inverse transforms $L^{-1}\{1/s\}=1$ and $L^{-1}\{s/(s^2+1)\}=\cos (t)$, the final result is $1-\cos (t)$.