`intx^2ln\ 4x\ dx`

We could let `u=x^2` or `u=ln\ 4x`.*.*

Considering the **priorities** given above, we
choose `u = ln\ 4x` and so `dv` will be the rest of the expression to be integrated `dv = x^2\ dx`.

With `u=ln\ 4x`, we have `du=dx/x`.

Integrating `dv = x^2\ dx` gives:

`v=intx^2dx=x^3/3`

Substituting in the Integration by Parts formula, we get:

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