Problem:Apply the inverse Laplace method to this expression: $F(p)=\frac{3p+7}{p^3}$
My Question: I know how to solve this and the right anwser to this problem. But I don't understand what rule is being used in the first step: $$3L^{-1}\{\frac{p}{p^3}\}+7L^{-1}\{\frac{1}{p^3}\}$$
I dont understand why we can write $L^{-1}\{\frac{p}{p^3}\}$ as $L^{-1}\{\frac{1}{2}\cdot\frac{2}{p^3}\}$ for the first transform. I mean why not $L^{-1}\{\frac{1}{2}\cdot\frac{2}{p^2}\}$ Why can we ignore p in numerator and write it as 1? What rule is being applied here? Can you, please, explain or give the name for this rule? I don't understand only this part of the problem... The anwser is $3t+\frac{7t^2}{2}$
Update: Due to confusion I repeat that
I am not asking about splitting a fraction I ask about transforming first fraction (which is $p/p^3$) and why we write it as $1/p^3$ and not $1/p^2$. I understand how and why we split the $(3p+7)/p^3$.
Update2: The part of the question is this expression: $3L^{-1}\{\frac{p}{p^3}\}$. The right approach is to write it as $3L^{-1}\{\frac{1}{p^3}\}$ and the multiply by $\frac{2}{2}$ to get $L^{-1}\{\frac{1}{2}\cdot\frac{2}{p^3}\}$, where $\frac{2}{p^3}$ is the table value. I dont understand why we write $p^3$ as denominator of this fraction and not $p^2$.
Update3: if $\frac{p}{p^3}=\frac{1}{p^2}$ why do we write it as $\frac{1}{p^3}$?
Laplace and Inverse Laplace Transforms are linear operators, such that $$L^{-1}(c)(a+b)=(c)L^{-1}(a)+(c)L^{-1}(b)$$ In your probelm, the fraction is merely split apart and evaluated.