Answer
$\dfrac{1}{2}e^{t}+e^{-t}+c$
Work Step by Step
Calculate the anti-derivative.
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and $ \int e^{-x} dx=-e^{-x}+c$
where $c$ is a constant of proportionality.
Then $\int (\dfrac{1}{2}e^t-e^{-t}) dt=\int \dfrac{1}{2}e^t dt-\int e^{-t} dt$
As we know $ \int e^{-x} dx=-e^{-x}+c$
or, $= \dfrac{1}{2}e^{t}+e^{-t}+c$
