If $ \mathrm{I}_{\mathrm{n}}=\int(\ln \mathrm{x})^{\mathrm{n}} \mathrm{dx} $, then $ \mathrm{I}_{\mathrm{n}}+\mathrm{nI}_{\mathrm{n}-1}= $

Question

If $ \mathrm{I}_{\mathrm{n}}=\int(\ln \mathrm{x})^{\mathrm{n}} \mathrm{dx} $, then $ \mathrm{I}_{\mathrm{n}}+\mathrm{nI}_{\mathrm{n}-1}= $, $ \frac{(\ln \mathrm{x})^{\mathrm{n}}}{\mathrm{x}}+\mathrm{C} $, $ \mathrm{x}(\ln \mathrm{x})^{\mathrm{n}-1}+\mathrm{C} $, $ \mathrm{x}(\ln \mathrm{x})^{\mathrm{n}}+\mathrm{C} $, None of these

MARKS App · Accepted Answer

Integrate by parts by taking ∫ ln x n dx as I n = ∫ ( l n x ) n .1. d x I n = x ln x n - ∫ x n ln x n - 1 x dx = x ln x n - n I n - 1 ⇒ I n + n I n - 1 = x ln x n

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