Q118 / 180 · KCET 2012 · Mathematics

If $\mathrm{n} \in \mathrm{N}$ and $\mathrm{I}_{\mathrm{n}}=\int(\log \mathrm{x})^{\mathrm{n}} \mathrm{dx}$, then $\mathrm{I}_{\mathrm{n}}+\mathrm{n} \mathrm{} \mathrm{I}_{\mathrm{n}-1}$ is equal to

  • A
    $\frac{(\log x)^{\mathrm{n}+1}}{\mathrm{n}+1}$
  • Correct answer
    B
    $x(\log x)^{n}+C$
  • C
    $(\log x)^{\mathrm{n}-1}$
  • D
    $\frac{(\log x)^{n}}{n}$

Solution

Here, $\mathrm{I}_{\mathrm{n}}=\int(\log \mathrm{x})^{\mathrm{n}} \mathrm{dx}$ On initegrating by parts, we get $$ \begin{aligned} &\mathrm{I}_{\mathrm{n}}=(\log \mathrm{x})^{\mathrm{n}} \cdot \mathrm{x}-\int \mathrm{x} \cdot \mathrm{n}(\log \mathrm{x})^{\mathrm{n}-1} \frac{1}{\mathrm{x}} \mathrm{dx} \\ &\mathrm{I}_{\mathrm{n}}=\mathrm{x}(\log \mathrm{x})^{\mathrm{n}} \mathrm{nI}_{\mathrm{n}-1} \\ &\therefore \quad \mathrm{I}_{\mathrm{n}}+\mathrm{nI}_{\mathrm{n}-1}=\mathrm{x}(\log \mathrm{x})^{\mathrm{n}}+\mathrm{C} \end{aligned} $$

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

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