Question

How do you integrate `int (3x+5)/(x-2)`?

Answer 1

`3x+11ln(abs(x-2))+C`

Explanation:

First simplify the integrand. This can be done by performing long division, or by my preferred method:

`int(3x+5)/(x-2)dx=int(3(x-2)+11)/(x-2)dx`

`color(white)(int(3x+5)/(x-2)dx)=int(3(x-2))/(x-2)dx+int11/(x-2)dx`

`color(white)(int(3x+5)/(x-2)dx)=3intdx+11intdx/(x-2)`

Both of these are simple integration problems. If you're stuck on the second one, try the substitution `u=x-2=>du=dx` and recall that `int(du)/u=ln(absu)+C`.

`color(white)(int(3x+5)/(x-2)dx)=3x+11ln(abs(x-2))+C`