Inverse laplace transform for $ \frac{s^2+1}{s^2(s+1)}$

Question

I used partial fractions so I equated the numerator $s^2+1=A(s+1) +Bs^2$ and I found $A$ and $B$ constants to be $A=1$ and $B=2$. When I inverse it I got the answer: $t +2\exp(-t)$ but the answer is $t + 2\exp(-t) -1$? From partial fractions I can only get two terms how did the inverse laplace $-1$ come about?

Answer 1

Hint

You need three fractions

$$ \frac{s^2+1}{s^2(s+1)}= \frac A s + \frac B {s^2}+\frac C {s+1}$$

$$ \begin{cases} B=1 \\ A+C=1\\ A+B=0 \end{cases} $$

Then, $$y(t)=A+Bt+Ce^{-t}$$