We find $du = 2\ln x \cdot \frac{1}{x} dx$ and $v = \frac{x^5}{5}$. Substituting these into the formula gives us:
$$\int_{1}^{2} x^{4}(\ln x)^{2} d x = \frac{x^5}{5}(\ln x)^2 - \int_{1}^{2} \frac{x^5}{5} \cdot 2\ln x \cdot \frac{1}{x} dx$$
which simplifies
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