Question

How do you find the indefinite integral of `int 1/(3-2x)`?

Answer 1

We can rewrite the integral as `int(-1/(2x - 3))dx`, which in turn can be rewritten as `int(-(2x - 3)^-1)dx`. We let `u = 2x - 3`, then `du= 2dx`, and `dx= (du)/2`

`=>int(-(u^-1))(du)/2`

`=>-1/2ln|u| + C`

`=>-1/2ln|2x- 3| + C`

Hopefully this helps!