$\mathrm{i)~\dfrac{s^2}{(s^2+a^2)(s^2+b^2)}}$ Let  $\mathrm{s^2=x}$ $\mathrm{\therefore\dfrac{s^2}{(s^2+a^2)(s^2+b^2)}=\dfrac{x}{(x+a^2)(x+b^2)}}$ Let  $\mathrm{\dfrac{x}{(x+a^2)(x+b^2)}=\dfrac{A}{(x+a^2)}+\dfrac{B}{(x+b^2)}}$ $\mathrm{\therefore x=A(x+b^2)+B(x+a^2)}$ Put $\mathrm{x=-a^2~~~~~~~~\therefore -a^2=A(-a^2+b^2)~~~~~~~~~\therefore A=\dfrac{a^2}{(a^2-b^2)}}$ Put $\mathrm{x=-b^2~~~~~~~~\therefore -b^2=B(-b^2+a^2)~~~~~~~~~\therefore B=\dfrac{-b^2}{(a^2-b^2)}}$ $\mathrm{\therefore\dfrac{s^2}{(s^2+a^2)(s^2+b^2)}=\dfrac{1}{a^2-b^2}\bigg[\dfrac{a^2}{s^2+a^2}-\dfrac{b^2}{s^2+b^2}\bigg]}$ $\mathrm{\therefore L^{-1}\bigg[\dfrac{s^2}{(s^2+a^2)(s^2+b^2)}\bigg]=\dfrac{1}{a^2-b^2}\bigg[L^{-1}\bigg(\dfrac{a^2}{s^2+a^2}\bigg)-L^{-1}\bigg(\dfrac{b^2}{s^2+b^2}\bigg)\bigg]}$ $\mathrm{=\dfrac{1}{a^2-b^2}\bigg[a^2\cdot L^{-1}\bigg(\dfrac{1}{s^2+a^2}\bigg)-b^2\cdot L^{-1}\bigg(\dfrac{1}{s^2+b^2}\bigg)\bigg]}$ $\mathrm{=\dfrac{1}{a^2-b^2}\bigg[\dfrac{a^2}{a}\cdot sin~at-\dfrac{b^2}{b}\cdot sin~bt\bigg]}$ $\mathrm{=\dfrac{1}{a^2-b^2}\Big(a~sin~at-b~sin~bt\Big)}$ $\mathrm{\therefore L^{-1}\bigg[\dfrac{s^2}{(s^2+a^2)(s^2+b^2)}\bigg]=\dfrac{1}{a^2-b^2}\Big(a~sin~at-b~sin~bt\Big)}$   $\mathrm{ii)~\dfrac{s+2}{s^2-4s+13}}$ Now, $\mathrm{L^{-1}\bigg[\dfrac{s+2}{s^2-4s+13}\bigg]=L^{-1}\bigg[\dfrac{s+2}{(s-2)^2+3^2}\bigg]}$ $\mathrm{=L^{-1}\bigg[\dfrac{(s-2)+4}{(s-2)^2+3^2}\bigg]=e^{2t}~L^{-1}\bigg[\dfrac{s+4}{s^2+3^2}\bigg]}$ $\mathrm{=e^{2t}~L^{-1}\bigg(\dfrac{s}{s^2+3^2}\bigg)+4e^{2t}~L^{-1}\bigg(\dfrac{1}{s^2+3^2}\bigg)}$ $\mathrm{=e^{2t}~cos~3t+\dfrac{4}{3}e^{2t}~sin~3t}$ $\mathrm{\therefore L^{-1}\bigg[\dfrac{s+2}{s^2-4s+13}\bigg]=e^{2t}~cos~3t+\dfrac{4}{3}e^{2t}~sin~3t}$