Yes.S=integration symbol

S (2/x) dx

do I move the x to the numerator and make it S(2x^-1) dx

Then ,2S(x^-1) dx

But when I do this and integrate it becomes x^0, which isn't working out too well.

EDIT:

I think I found what to do,

If I have

S(2/x) dx=

2S(1/x) dx=

2lnx, with the x being absolute value.

Am I on the right track?

Would S (3/x) dx be:

3S(1/x) dx=

3 lnx ?

S(4/x) dx=

4S(1/x)=

4lnx?

and so on?

RonL