# Find the antiderivative of e^{-0.5t} + 1

## Question:

Find the antiderivative of {eq}e^{-0.5t} + 1 {/eq}

## Antiderivative

The antiderivative of any function is also known as the indefinite integral of the function.

To solve the indefinite integral, we'll use the integral sum rule, which is written as {eq}\int (f(x)+ g(x))dx=\int f(x)dx + \int g(x)dx {/eq}

Next, we use the common integral formulas:

{eq}\int e^{ax}dx=\dfrac{e^{ax}}{a}\\ \int dx=x {/eq}

We need to find out the antiderivative of {eq}e^{-0.5t} + 1 {/eq}

Hence, we need to compute {eq}\int (e^{-0.5t} + 1) dt {/eq}

Now:

{eq}\begin{align} \int (e^{-0.5t} + 1) dt &=\int e^{-0.5t}dt+\int 1dt \ & \left [ \because \int (f(x)+ g(x))dx=\int f(x)dx + \int g(x)dx \right ]\\ &=\dfrac{e^{-0.5t}}{-0.5}+t+C \ & \left [ \because \int e^{ax}dx=\dfrac{e^{ax}}{a}, \int dx=x \right ]\\ &=-\dfrac{e^{-0.5t}}{0.5}+t+C\\ &=-\dfrac{e^{-0.5t}}{\frac{5}{10}}+t+C\\ &=-2e^{-0.5t}+t+C \end{align} {/eq}

where {eq}C {/eq} is constant of integration.

{eq}\color{blue}{\boxed{\int (e^{-0.5t} + 1) dt=-2e^{-0.5t}+t+C}} {/eq}