Q112 / 160 · AP EAMCET 2023 (15 May Shift 2) · Mathematics

$\int x^3(\log x)^2 d x=$

  • A
    $(\log x)^2 \frac{x^4}{4}+\frac{1}{2}\left[(\log x) \frac{x^4}{4}+\frac{x^4}{16}\right]+C$
  • B
    $(\log x)^2 \frac{x^4}{4}-\frac{1}{2}\left[(\log x) \frac{x^4}{4}+\frac{x^4}{16}\right]+C$
  • Correct answer
    C
    $(\log x)^2 \frac{x^4}{4}-\frac{1}{2}\left[(\log x) \frac{x^4}{4}-\frac{x^4}{16}\right]+C$
  • D
    $(\log x)^2 \frac{x^4}{4}+\frac{1}{2}\left[(\log x) \frac{x^4}{4}-\frac{x^4}{16}\right]+C$

Solution

$\begin{aligned} & \text {Let } I=\int x^3(\log x)^2 d x \\ & =(\log x)^2 \cdot \frac{x^4}{4}-\int \frac{2}{4}(\log x) \cdot x^3 d x \\ & =(\log x)^2 \cdot \frac{x^4}{4}-\frac{1}{2}\left[(\log x) \cdot \frac{x^4}{4}-\int \frac{x^3}{4} d x\right] \\ & =(\log x)^2 \frac{x^4}{4}-\frac{1}{2}\left[(\log x) \frac{x^4}{4}-\frac{x^4}{16}\right]+c\end{aligned}$

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

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