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Question: If \(I_{n} = \int_{}^{}{(\log x)^{n}dx,}\) then \(I_{n} + nI_{n - 1} =\)...

If In=∫(log⁡x)ndx,I_{n} = \int_{}^{}{(\log x)^{n}dx,} then In+nIn−1=I_{n} + nI_{n - 1} =

A

x(log⁡x)nx(\log x)^{n}

B

(xlog⁡x)n(x\log x)^{n}

C

(log⁡x)n−1(\log x)^{n - 1}

D

n(log⁡x)nn(\log x)^{n}

Answer

x(log⁡x)nx(\log x)^{n}

Explanation

Solution

In=∫(log⁡x)ndxI_{n} = \int_{}^{}{(\log x)^{n}dx}, ∴In−1=∫(log⁡x)n−1dx\therefore I_{n - 1} = \int_{}^{}{(\log x)^{n - 1}}dx

Now In=∫(log⁡x)n.1dx=(log⁡x)n.x−n∫(log⁡x)n−1.1x.xdx=(log⁡x)n.x−n∫(log⁡x)n−1dxI_{n} = \int_{}^{}{(\log x)^{n}.1dx} = (\log x)^{n}.x - n\int_{}^{}{(\log x)^{n - 1}.\frac{1}{x}.xdx} = (\log x)^{n}.x - n\int_{}^{}{(\log x)^{n - 1}dx}

In=x(log⁡x)n−nIn−1I_{n} = x(\log x)^{n} - nI_{n - 1} ∴In+nIn−1=x(log⁡x)n\therefore I_{n} + nI_{n - 1} = x(\log x)^{n}