If $I_{n}=\int(\ln x)^n d x$ then $I_n+n I_{n-1}$ is equal to
Answer & explanation
Correct answer: option 2
$I_{n}=\int(\ln x)^n d x$
$I_n=x(\ln x)^n-n \int(\ln x)^{n-1} d x$
$I_n=x(\ln x)^n-n I_{n-1}$
$I_n+n I_{n-1}=x(\ln x)^n$
Hence (2) is the correct answer.