Q89 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics

$\int x^3 \log x\,dx = $

  • Correct answer
    A
    $\dfrac{x^4}{16}[4\log x - 1] + c$
  • B
    $\dfrac{x^4}{16}[4\log x + 1] + c$
  • C
    $\dfrac{x^4}{16}[-4\log x - 1] + c$
  • D
    $\dfrac{x^4}{16}[-4\log x + 1] + c$

Solution

Using integration by parts, $\int u v \, dx = u \int v \, dx - \int \left( \dfrac{du}{dx} \int v \, dx \right) dx$

Let $u = \log x$ and $v = x^3$

$\int x^3 \log x \, dx = (\log x) \left( \dfrac{x^4}{4} \right) - \int \left( \dfrac{1}{x} \left( \dfrac{x^4}{4} \right) \right) dx$

$= \dfrac{x^4}{4} \log x - \int \dfrac{x^3}{4} \, dx$

$= \dfrac{x^4}{4} \log x - \dfrac{x^4}{16} + c$

$= \dfrac{x^4}{16}[4\log x - 1] + c$

Answer: $\dfrac{x^4}{16}[4\log x - 1] + c$

Hence, the correct answer is option: 'A'.

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