# Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace tranform of the function below.

Jason Farmer 2021-09-24 Answered
Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace tranform of the function below.
$$\displaystyle{\left({1}+{e}^{{-{4}{t}}}\right)}^{{2}}$$

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## Expert Answer

Jozlyn
Answered 2021-09-25 Author has 13136 answers

We have to calculate the laplace transform of $$\displaystyle{\left({1}+{e}^{{-{4}{t}}}\right)}^{{2}}$$
Now $$\displaystyle{\left({1}+{e}^{{-{4}{t}}}\right)}^{{2}}$$
$$\displaystyle={\left({1}\right)}^{{2}}+{\left({e}^{{-{4}{t}}}\right)}^{{2}}+{2}\cdot{1}\cdot{e}^{{-{4}{t}}}$$
$$\displaystyle={1}+{e}^{{-{8}{t}}}+{2}{e}^{{-{4}{t}}}$$
now $$L\left\{(1+e^{-4t})^2\right\}$$
$$=L\left\{1+e^{-8t}+2e^{-4t}\right\}$$
$$=L\left\{1\right\}+L\left\{e^{-8t}\right\}+L\left\{2e^{-4t}\right\}$$
$$=L\left\{1\right\}+L\left\{e^{-8t}\right\}+2L\left\{e^{-4t}\right\}$$
$$=\frac{1}{p}+\frac{1}{p+8}+2\frac{1}{p+4} \ \ \left\{\text{as } L\left\{e^{-at}\right\}=\frac{1}{p+a}\right\}$$
$$\therefore L\left\{(1+e^{-4t})^2\right\}=\frac{1}{p}+\frac{1}{p+8}+\frac{2}{p+4}$$

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