Use integration by parts to establish a reduction formula for the integral.
-$\int ( \ln a x ) ^ { n } d x$
A) $\int ( \ln a x ) ^ { \mathrm { n } } \mathrm { dx } = a x ( \ln a x ) ^ { \mathrm { n } } - a n \int ( \ln a x ) ^ { \mathrm { n } - 1 } \mathrm { dx }$
B) $\int ( \ln a x ) ^ { n } d x = x ( \ln a x ) ^ { n } - n \int ( \ln a x ) ^ { n - 1 } d x$
C) $\int ( \ln a x ) ^ { n } d x = \frac { x ( \ln a x ) ^ { n } } { n } + \frac { n } { a } \int ( \ln a x ) ^ { n - 1 } d x$
D) $\int ( \ln a x ) ^ { n } d x = \frac { x ( \ln a x ) ^ { n } } { n } - \frac { n } { a } \int ( \ln a x ) ^ { n - 2 } d x$

Question

Quizplus · Accepted Answer

Let $u=\ln(ax)^n$ and $dv=dx$. Then $du=\frac{n(\ln(ax))^{n-1}}{x}dx$ and $v=x$. Using integration by parts, we get \begin{align*}
\int(\ln(ax))^ndx&=x(\ln(ax))^n-\int \frac{n(\ln(ax))^{n-1}}{x}x\mathrm dx\
&=x(\ln(ax))^n-n\int(\ln(ax))^{n-1}\mathrm dx\
&=x(\ln(ax))^n-n\int\frac{(\ln(ax))^n}{\ln(ax)}\mathrm d(\ln(ax))\
&=x(\ln(ax))^n-n\int(\ln(ax))^n\mathrm d(\ln(ax))\
&=x(\ln(ax))^n-n\cdot\frac{1}{n+1}(\ln(ax))^{n+1}+C\
&=x(\ln(ax))^n-\frac{n}{n+1}(\ln(ax))^{n+1}+C.
\end{align*} Therefore, the answer is $\boxed{	ext{(B)}}$.

Use Integration by Parts to Establish a Reduction Formula for the Integral