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∫14x3log⁡x dx\int_1^4 x^3\log x\, dx is equal to:

Solution

✅ Correct Option: 3

Integrating by parts with log⁡x\log x as the first function gives x44log⁡x−x416\frac{x^4}{4}\log x-\frac{x^4}{16}.

At x=4x=4: 64log⁡4−1664\log4-16; at x=1x=1: −116-\frac{1}{16}.

The difference is 128log⁡2−25516128\log2-\frac{255}{16}, using 64log⁡4=128log⁡264\log4=128\log2.

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