## Calculus 8th Edition

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