Answer
$\displaystyle \frac{32}{5}\ln^2(2) - \frac{64}{25}\ln(2) + \frac{62}{125} ≈ 1.796442$
Work Step by Step
Integration by parts formula: $\displaystyle \int fg' $ $\displaystyle dx = fg-\int f'g$ $dx$
$\displaystyle \int_1^2x^4[\ln(x)]^2dx$
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Let $\displaystyle f = \ln^2(x)$ and $\displaystyle g' = x^4$
thus $\displaystyle f' = 2\ln(x)(\frac{1}{x})$ and $\displaystyle g = \frac{x^5}{5}$
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$\displaystyle \int_1^2x^4[\ln(x)]^2dx = [\frac{x^5}{5}\ln^2(x)]^2_1 - \int_1^2\frac{x^5}{5}[2\ln(x)(\frac{1}{x})])dx$
$\displaystyle [\frac{32}{5}\ln^2(2) - \frac{1}{5}\ln^2(1)] - \int_1^2\frac{2x^4}{5}\ln(x)dx$
$\displaystyle [\frac{32}{5}\ln^2(2) - 0] - ([\frac{2x^5}{25}\ln(x)]_1^2)-\int_1^2\frac{2x^5}{25}(\frac{1}{x})dx)$
$\displaystyle \frac{32}{5}\ln^2(2) - ([\frac{64}{25}\ln(2) - \frac{2}{25}\ln(1)] - \frac{2}{25}\int_1^2x^4dx)$
$\displaystyle \frac{32}{5}\ln^2(2) - ([\frac{64}{25}\ln(2) - 0] - \frac{2}{25}[\frac{x^5}{5}]_1^2)$
$\displaystyle \frac{32}{5}\ln^2(2) - [\frac{64}{25}\ln(2) - \frac{2}{25}(\frac{32}{5}-\frac{1}{5})]$
$\displaystyle \frac{32}{5}\ln^2(2) - [\frac{64}{25}\ln(2) - \frac{2}{25}(\frac{31}{5})]$
$\displaystyle \frac{32}{5}\ln^2(2) - [\frac{64}{25}\ln(2) - \frac{62}{5}]$
$\displaystyle \frac{32}{5}\ln^2(2) - \frac{64}{25}\ln(2) + \frac{62}{125} ≈ 1.796442$
