Question

How do you integrate `lnx / x`?

Answer 1

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`