The Laplace transform of $$f\left( t \right)$$ is $${1 \over {{s^2}\left( {s + 1} \right)}}.$$
The function

The inverse Laplace transform of $${1 \over {\left( {{s^2} + s} \right)}}$$ is

If $$F(s)$$ is the Laplace transform of the function $$f(t)$$ then Laplace transform of $$\int\limits_0^t {f\left( x \right)dx} $$ is

Laplace transform of $${\left( {a + bt} \right)^2}$$ where $$'a'$$ and $$'b'$$ are constants is given by: