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Question
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Solution
\[\int {x^3}_{II} \cdot \left( \log_I x \right)^2 \cdot dx\]
\[ = \left( \log x^2 \right)\int x^3 dx - \int\frac{2 \log x}{x} \times \frac{x^4}{4} \text{ dx} \]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2}\int \log_I x \cdot {x^3}_{II} \text{ dx }\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2}\left[ \log x\int x^3 dx - \int\left\{ \frac{d}{dx}\left( \log x \right)\int x^3 dx \right\}dx \right]\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2} \left[ \log x \cdot \frac{x^4}{4} - \int\frac{1}{x} \times \frac{x^4}{4}dx \right]\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2} \left[ \log x \cdot \frac{x^4}{4} - \frac{1}{4}\int x^3 dx \right]\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2} \left[ \log x \cdot \frac{x^4}{4} - \frac{x^4}{16} \right] + C\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{\log x \cdot x^4}{8} + \frac{x^4}{32} + C\]
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