Integration $1/e^{(2-3x)} dx$

Question

One question in integration is to integrate $1/e^{(2-3x)} dx$ the provided answer is to let $e^{-(2-3x)} = e^{(3x-2)}$ then integrate it to be $1/3 e^{3x-2}$ the question is why we can't use $u = (2-3x)$ then $du= -3$ so it should be $-1/3 (1/u) du = -1/3 \ln|u| $ instead of the provided answer

Thanks

Answer 1

Because your denominator is $e^u$, not $u$, so you can't integrate to $\ln |u|$

Answer 2

You can certainly obtain the correct result using $u = 2-3x$ and $du = -3dx$, however you would need to use two substitutions, and you substituted the quantity back into the integral incorrectly.

Upon substitution, you would get $$\int\frac1{e^{2-3x}}dx \to \int\frac{-1}{3e^u}du = -\frac{1}{3}\int e^{-u}du$$

Then, you would need to substitute $v = -u$ and $dv = -du$ to get $$\frac1{3}\int e^v dv = \frac13e^v + C = \frac13e^{-u} + C = \frac13e^{3x-2} + C$$

As you can see, this still gives you the desired result, but it was a far more roundabout way to do it. Of course, $\int e^{-u} du$ is a fairly trivial integral to do in your head, so most people don't need to do a second level of substitution, but behind the scenes, that is what's happening.