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

Question

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!

Answer 1

if $I_m$ is the integral of $\log^m x$ then integration by parts gives: $$ I_{m} = x \log^{m} x - m I_{m-1} $$

by repeated application of this we obtain:

$$ I_m = x( \log^m x - m\log^{m-1} x + m(m-1) \log^{m-2} x - ... +(-1)^m) $$ which may be written as: $$ I_m = x\sum_{k=0}^m(-1)^{m-k} \frac{m!}{k!} \log^k x $$

Answer 2

$$ \int \ln^{n} x dx = x \ln^{n}x - n\int\ln^{n-1}x dx + C $$

Please see:

https://www.youtube.com/watch?v=xE0Pp4I7PiA

Answer 3

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: $$I_n=x\ln^n x -n\int x \ln ^{n-1}x~ dx\Rightarrow I_n=x \ln^n x-n I_{n-1}, ~~~I_0=x.$$