How do you integrate #lnx / x#?

1 Answer
Apr 30, 2016

Answer:

Use a #u#-substitution to get #(lnx)^2/2+C#.

Explanation:

At first glance, this integral looks a little confusing because we have a function divided by another function (and those tend to be difficult to work with). But, after rewriting #intlnx/xdx# as #int1/xlnxdx#, we can see something interesting: we have #lnx# and its derivative, #1/x#, in the same integral, making it a textbook case of a #u#-substitution:
Let #color(blue)u=color(blue)lnx->(du)/dx=1/x->color(red)(du)=color(red)(1/xdx)#

Thus, the integral #intcolor(red)(1/x)color(blue)(lnx)color(red)(dx)# becomes:
#intcolor(blue)(u)color(red)(du)#

Now, isn't this much easier? Using the reverse power rule, the integral evaluates to #u^2/2+C#. Because #u=lnx#, we can say:
#intlnx/xdx=(lnx)^2/2+C#