Solveeit Logo
Login

Question

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

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

A

x (log x)n

B

(x log x)n

C

(log x) n –1

D

n(logx) n

Answer

x (log x)n

Explanation

Solution

In = ∫(log⁡x)n.1dx\int_{}^{}{(\log x)_{}^{n}}.1dx

= = (log⁡x)nx−nIn−1(\log x)^{n}x - nI_{n - 1}

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