Q. If $I_{n} = \int_{}^{}\left( \log x \right)^{n}dx$, then $I_{n} + nI_{n - 1}$ is equal to
A) $\left( x\log x \right)^{n}$
B) $x\left( \log x \right)^{n}$
C) $n\left( \log x \right)^{n}$
D) $\left( \log x \right)^{n - 1}$
Explanation
(b)
$I_{n} = \int_{}^{}\left( \log x \right)^{n}dx$
$= \left( \log x \right)^{n}.x - \int_{}^{}x.n\left( \log x \right)^{n - 1}\frac{1}{x}dx\$
$\Rightarrow \ I_{n} = x\left( \log x \right)^{n} - nI_{n - 1}$I
$\Rightarrow \ I_{n} + nI_{n - 1} = x\left( \log x \right)^{n}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.