Question

How do you find the integral of `int z/e^(3z) dz`?

Answer 1

`-z/3 e^{-3z}-1/9 e^{-3z}+C`

Explanation:

Using integration by parts (with `z` as the "first function" and `e^{-3z}` as the "second function"):

`int z e^{-3z} dz = z times int e^{-3z}dz - int {(d/dz z) int e^{-3z}dz } dz`
`= z times (-1/3 e^{-3z})- int {1 times (-1/3 e^{-3z})}dz`
`=-z/3 e^{-3z} + 1/3 int e^{-3z} dz = -z/3 e^{-3z}-1/9 e^{-3z}+C`