Question

What is the antiderivative of `ln(x)^2`?

Answer 1

`int (lnx)^2dx = x(lnx)^2 +2x +C`

Explanation:

You should learn the IBP formula:
`int u(dv)/dxdx=uv - int v (du)/dxdx`

So essentially we are looking for one function that simplifies when it is differentiated, and one that simplifies when integrated (or at least is integrable).

We also need to know that `int lnxdx=xlnx-x` (either learn or use IBP):

Let `{(u=lnx, => ,(du)/dx=1/x),((dv)/dx=lnx,=>,v =xlnx-x ):}`

Then IBP gives;
`int lnx lnx dx=lnx(xlnx-x) - int (xlnx-x)1/xdx`
`:. int (lnx)^2 dx = x(lnx)^2 - xlnx - int (lnx-1)dx`
`:. int (lnx)^2 dx = x(lnx)^2 - xlnx - (xlnx-x-x)`
`:. int (lnx)^2 dx = x(lnx)^2 - xlnx + xlnx +2x`

And so we have:
`:. int (lnx)^2dx = x(lnx)^2 +2x +C`