# Find the laplace transform of the following 1) SHIFTING f(t)=(1+te^{-t})^3

Find the laplace transform of the following
1) SHIFTING
$f\left(t\right)=\left(1+t{e}^{-t}{\right)}^{3}$
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Step 1
Simplify:
$f\left(t\right)=\left(1+t{e}^{-t}{\right)}^{3}$
$=1+3t{e}^{-t}+3{t}^{2}{e}^{-2t}+{t}^{3}{e}^{-3t}$
Step 2
Calculate Laplace transform of ${e}^{-at}$:
${\int }_{0}^{\mathrm{\infty }}{e}^{-at}{e}^{-st}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+a\right)t}dt$
$=\frac{1}{\left(s+a\right)}$
Step 3 Thus, Laplace transform of ${t}^{n}{e}^{-at}$ is:
$\left(-1{\right)}^{n}\frac{{d}^{n}}{d{s}^{n}}\left(\frac{1}{\left(s+a\right)}\right)=\left(-1{\right)}^{n}\frac{{d}^{n}}{d{s}^{n}}\left(s+a{\right)}^{-1}$
$=\frac{\left(n!\right)}{\left(s+a{\right)}^{n+1}}$
Step 4 Therefore Laplace transform of
$\frac{\left(0!\right)}{\left(s+0{\right)}^{0+1}}+\frac{\left(31!\right)}{\left(s+1{\right)}^{1+1}}+\frac{\left(32!\right)}{\left(s+2{\right)}^{2+1}}+\frac{\left(3!\right)}{\left(s+3{\right)}^{3+1}}=\frac{1}{s}+\frac{3}{\left(s+1{\right)}^{2}}+\frac{6}{\left(s+2{\right)}^{3}}+\frac{6}{\left(s+3{\right)}^{4}}$
Step 5
Therefore Laplace transform of f(t) is $\frac{1}{s}+\frac{3}{\left(s+1{\right)}^{2}}+\frac{6}{\left(s+2{\right)}^{3}}+\frac{6}{\left(s+3{\right)}^{4}}$
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