Question

How do you evaluate `int (1/(3-5x)^2)dx` for [1, 2]?

Answer 1

Use substitution with `u = 3-5x`

Explanation:

`int_1^2 1/(3-5x)^2dx`

Let `u = 3-5x`, so that `du = -5 dx` and
when `x=1`, we get `u = -2`
when `x=2`, we get `u = -7`

The integral becomes

`-1/5 int_-2^-7 1/u^2 du = -1/5 [-1/u]_-2^-7`

`= -1/5[-1/(-7) - ( - 1/(-2)) = -1/5[1/7 - 1/2]`

`= -1/5[-5/14] = 1/14.`