I'm trying to find the inverse Laplace transform of:
$$\frac{as+b}{s(cs+d)+g}\tag1$$
First of all I can expand the fraction:
$$\frac{as+b}{s(cs+d)+g}=a\cdot\frac{ s}{s(cs+d)+g}+b\cdot\frac{1}{s(cs+d)+g}\tag2$$
And from now on I've no idea
Background: this inverse Laplace transform is important in solving a second order DE.
$$s(cs+d)+g=cs^2+ds+g=\left(\sqrt{c}s+\frac{d}{2\sqrt{c}}\right)^2+\left(g-\frac{d^2}{4c}\right)$$ so we can say: $$\frac{as+b}{s(cs+d)+g}=\frac{as+b}{\left(\sqrt{c}s+\frac{d}{2\sqrt{c}}\right)^2+\left(g-\frac{d^2}{4c}\right)}$$ or we could rewrite it as: $$\frac{as+b}{\left(\sqrt{c}s+\frac{d}{2\sqrt{c}}\right)^2-\left(\frac{d^2}{4c}-g\right)}=\frac{as+b}{\left(\sqrt{c}s+\frac{d}{2\sqrt{c}}\right)^2-\sqrt{\frac{d^2}{4c}-g}^2}$$
You could try and make a substitution from this i.e. let $u=\sqrt{c}s+\frac{d}{2\sqrt{c}}$ and then use partial fractions, calculate the inverse Laplace transform then backsubstitute