Find $\int\ln^{n} x$ . 
Find $\int\ln^{n} x$ .

My observation
$\int\ln^{1} x= x(\ln x- 1)+ constant$
$\int\ln^{2} x= x(\ln^{2} x- 2\ln x+ 2)+ constant$
$\int\ln^{3} x= x(\ln^{3} x- 3\ln^{2} x+ 6\ln x- 6)+ constant$
$\int\ln^{4} x= x(\ln^{4} x- 4\ln^{3} x+ 12\ln^{2} x- 24\ln x+ 24)+ constant$
$$\ddots$$
We have
$$\ddots$$
$\frac{{\rm d}}{{\rm d}\ln x} (\ln^{4} x- 4\ln^{3} x+ 12\ln^{2} x- 24\ln x+ 24)= 4(\ln^{3} x- 3\ln^{2} x+ 6\ln x- 6)$
$\frac{{\rm d}}{{\rm d}\ln x} (\ln^{3} x- 3\ln^{2} x+ 6\ln x- 6)= 3(\ln^{2} x- 2\ln x+ 2)$
$\frac{{\rm d}}{{\rm d}\ln x} (\ln^{2} x- 2\ln x+ 2)= 2(\ln x- 1)$
I used these to prep for my tests, thanks!

Let $x=e^t$, Then $$I_n=\int \ln^n x~ dx= \int t^n e^t dt~ \mbox{(Int. by parts)}~= t^n e^t -n\int t^{n-1} e^t~ dt = t^n e^t -n I_{n-1}.$$ This is the reduction formula, using which
repeatedly on get the desired result. Lastly, we have $I_0= t$ Or alternatively: $$I_n=x\ln^n x -n\int x \ln ^{n-1}x~ dx\Rightarrow I_n=x \ln^n x-n I_{n-1}, ~~~I_0=x.$$
