I am trying to find the answer to the Laplace Transformation where $f(t) = {(2t+1)}^3$
From technique and procedure, we use Integration by Parts twice, with
$$uv - \int_0^{\infty}{vdu}$$ $$mn - \int_0^{\infty}{ndm}$$
Substituting $f(t)$ we have
$$\left[-\frac{{(2t+1)}^3e^{-st}}{s}\right]_{0}^{\infty} +\frac{6(2t+1)}{s}\int_0^{\infty}{e^{-st}{(2t+1)^2}}dt$$
$$(-1)\left[\frac{{(2t+1)}^3e^{-st}}{s} +\frac{6(2t+1)^2}{s^2}\right]_{0}^{\infty} +\frac{24(2t+1)e^{-st}}{s^2}\int_0^{\infty}{e^{-st}{(2t+1)}}dt$$
Is this correct so far, and I just take the last integrand? And make sure to carry the $(-1)$ down.
Edit: what i think is that $\frac{24(2t+1)e^{-st}}{s^2}$ leans more towards $$\frac{48te^{-st}}{s^3}\int_0^{\infty}{e^{-st}}dt$$
I am using the reference Limits in Laplace Transformation2
without substituting the $a$. If I could have guidence to that for this particular case, it would be helpful. Perhaps I could apply that to more problems (as for the case in the link).
I would just expand the cube and evaluate the 4 terms. $$ (2t+1)^3=1 + 6 t + 12 t^2 + 8 t³ $$ $$ \mathcal{L}_t\left((2 t+1)^3\right)\left[s \right]=\frac{48}{s^4}+\frac{24}{s^3}+\frac{6}{s^2}+\frac{1}{s} $$