Question

What is `f(x) = int 1/(x-3)-1/(x-2) dx` if `f(-1)=6`?

Answer 1

`f(x) = ln|((x - 3)/(x - 2))| + 6 - ln(4/3)`

Explanation:

We use the commonly used result `int (1/x) dx = ln|x|` to solve.

`f(x) = ln|x - 3| - ln|x - 2| + C`

`f(x) = ln|(x- 3)/(x - 2)| + C`

Now we determine the value of the constant of integration.

`6 = ln|(-1 - 3)/(-1 - 2)| + C`

`6 = ln|4/3| + C`

`C = 6 - ln(4/3)`

So the function is

`f(x) = ln|((x - 3)/(x - 2))| + 6 - ln(4/3)`

Hopefully this helps!