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