Question about inverse Laplace transform

Question

I have here a probably a very easy inverse Laplace transform.

I just can not figure out why $$ L^{-1}( \frac{1}{s^2} \frac{s-a}{s+a}) =-t + \frac2a- \frac2a e^{-at} $$

Is there a conversion I don't see? I appreciate any help with !

Answer 1

$$f(s)= \frac{1}{s^2} \frac{s-a}{s+a} $$ $$f(s)= \dfrac 1 s \left(\dfrac{1}{s} \dfrac{s-a}{s+a} \right) $$ $$f(s)= \dfrac 1 s \left(\dfrac{A}{s}+ \dfrac{B}{s+a} \right) $$ $$f(s)= \dfrac 1 s \left ( \dfrac{(A+B)s+Aa}{s(s+a)}\right) $$ Where $A+B=1$ and $A=-1 \implies B=2$ so that we have: $$f(s)= \dfrac 1 s \left(-\dfrac{1}{s}+ \dfrac{2}{s+a}\right) $$ $$f(s)= -\dfrac{1}{s^2}+ \dfrac{2}{s(s+a)} $$ $$f(s)= -\dfrac{1}{s^2}+ \dfrac 2 a \left(\dfrac{1}{s}-\dfrac {1}{s+a} \right) $$ Apply inverse Laplace transform now. $$f(t)= -t+ \dfrac 2 a \left(1-e^{-at} \right) $$