Question

How do you integrate `(lnx)^2`?

Answer 1

First, note that `\int ln(x) dx` can be done by parts by letting `u=ln(x), dv=dx` so that `du=dx/x` and `v=x`, leading to `\int ln(x) dx=x ln(x)-\int dx=x ln(x)-x+C`.

Now, for `\int (ln(x))^2 dx`, use integration-by-parts again with `u=ln(x)` and `dv=ln(x) dx` so that `du = dx/x` and `v=x ln(x)-x`. Then `\int(\ln(x))^2 dx=x(ln(x))^2-x ln(x)-\int (ln(x)-1)dx= x(ln(x))^2-x ln(x)+x-(x ln(x)-x)`

`=x(ln(x))^2-2xln(x)+2x+C`.