Chapter 6
Antidifferentiation
6.1 Finding Antiderivatives
6.1.2 Antiderivatives Involving Exponential and
Logarithmic Functions
Find an antiderivative for `e^(5t).`
Solution Since `d/(dt)e^t=e^t,` we know by the Chain Rule that `d/(dt)e^(5t)=5e^(5t).` Thus an antiderivative for `e^(5t)` is `1/5e^(5t).`
Find an antiderivative for `1/(3x+1).`
Solution The reciprocal function is the one exception to our general rule for finding antiderivatives of power functions:
`x^(r+1)/(r+1)` is an antiderivative for `x^r,` except when `r=-1.`
However, we know an antiderivative for `1/x`:
Thus, by the Chain Rule,
Using the Constant Multiple Rule we see that
So, `1/3 ln(3x+1)` is an antiderivative for `1/(3x+1).`
Next we note that the Sum Rule for differentiation enables us to calculate antitderivatives term by term.
Find an antiderivative for `3x^5+sin(x/2)-e^(-x).`
Solution Taking each of the terms separately, we calculate the following:
An antiderivative for `3x^5` is `1/2 x^6.`
An antiderivative for `sin(x/2)` is `-2 cos(x/2).`
An antiderivative for ` -e^(-x)` is ` e^(-x).`
Putting this together, we have the antiderivative that we want:

`d/(dx)[1/2 x^6-2 cos(x/2)+e^(-x)]` | `=d/(dx)(1/2 x^6)+d/(dx)[-2 cos(x/2)]+d/(dx)e^(-x)` |
`=3x^5+sin(x/2)-e^(-x)`. |