What is #int_3^oo 3/x -2/(x-2)dx#?

1 Answer
Jun 12, 2017

Well, think about what this integral is asking you to do. You are starting at the left at #x = 3# and integrating rightwards towards infinity... and the function decreases towards a horizontal asymptote...

If you integrate #3/x - 2/(x-2)# from #3# to #oo#, you've attempted to integrate in a half-open interval, so the integral is not finite.

The graph looks like this:

graph{3/x - 2/(x-2) [-9.19, 19.29, -3.32, 10.91]}

and Wolfram Alpha gives...


#int 3/x - 2/(x-2)dx = 3ln|x| - 2ln|x-2|#

Evaluating from #3# to #oo# gives:

#= lim_(x->oo) [3ln|x| - 2ln|x-2|] - [3ln|3| - cancel(2ln|3-2|)^(0)]#

#= oo - oo' - 3ln|3|#

#=># #color(blue)("DNE")#