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Question: If I<sub>n</sub> = \(\int_{}^{}{(\log x)^{n}}\)dx, then I<sub>n</sub> + nI<sub>n–1</sub> is equal to...

If In = ∫(log⁡x)n\int_{}^{}{(\log x)^{n}}dx, then In + nIn–1 is equal to –

A

x(log x)n

B

(x log x)n

C

(log x)n – 1

D

n(log x)n

Answer

x(log x)n

Explanation

Solution

In = ∫(log⁡x)ndx\int_{}^{}{(\log x)^{n}dx} …(i)

\ In–1 = ∫(log⁡x)n−1dx\int_{}^{}{(\log x)^{n - 1}dx} …(ii)

Now, In = . 1 dx

= (log x)n x – n 1x\frac{1}{x}x dx

= x(log x)n – n∫(log⁡x)n−1\int_{}^{}{(\log x)^{n - 1}}dx

In = x (log x)n – n In – 1 \ In + nIn–1 = x(log x)n