`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:
Substituting in the Integration by Parts formula, we get:
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