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

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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Jozlyn

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 $$L\left\{(1+e^{-4t})^2\right\}$$
$$=L\left\{1+e^{-8t}+2e^{-4t}\right\}$$
$$=L\left\{1\right\}+L\left\{e^{-8t}\right\}+L\left\{2e^{-4t}\right\}$$
$$=L\left\{1\right\}+L\left\{e^{-8t}\right\}+2L\left\{e^{-4t}\right\}$$
$$=\frac{1}{p}+\frac{1}{p+8}+2\frac{1}{p+4} \ \ \left\{\text{as } L\left\{e^{-at}\right\}=\frac{1}{p+a}\right\}$$
$$\therefore L\left\{(1+e^{-4t})^2\right\}=\frac{1}{p}+\frac{1}{p+8}+\frac{2}{p+4}$$