Chapter 6
Antidifferentiation





6.1 Finding Antiderivatives

6.1.2 Antiderivatives Involving Exponential and
         Logarithmic Functions


Example 4

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).`

Checkpoint 4Checkpoint 4

Example 5

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`:

`d/(dx) ln(x)=1/x`.

Thus, by the Chain Rule,

`d/(dx) ln(3x+1)=3/(3x+1)`.

Using the Constant Multiple Rule we see that

`d/(dx) 1/3 ln(3x+1)=1/3 times d/(dx) ln(3x+1) = 1/3 times 3 times 1/(3x+1)=1/(3x+1)`.

So, `1/3 ln(3x+1)` is an antiderivative for `1/(3x+1).`

Checkpoint 5Checkpoint 5

Next we note that the Sum Rule for differentiation enables us to calculate antitderivatives term by term.

Example 6

Find an antiderivative for `3x^5+sin(x/2)-e^(-x).`

Solution   Taking each of the terms separately, we calculate the following:

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)`.

Checkpoint 6Checkpoint 6

Checkpoint 7Checkpoint 7

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