Using integration by parts to prove a reduction formula

Question

I could use some insights on using integration by parts to prove the following reduction:

$\int ln(x)^n dx = x ln(x)^n - n \int ln(x)^{n-1}dx$

I have gone as far as showing that

$\int ln(x)^n dx = x ln(x)^n - n \int ln(x)^{n-1}dx - \int ln(x)^{n-1} - \left[xln(x)^{n-1} - (n-1) \int ln(x)^{n-2}dx\right]$

Now I know that if the reduction is true, then $\int ln(x)^{n-1} = xln(x)^{n-1} - (n-1) \int ln(x)^{n-2}dx$ so they will cancel out. However I cannot use the equality I am trying to prove in the proof itself, right?

Any help would be much appreciated!

Answer 1

I think you have meant $\displaystyle \int (\ln x)^ndx=I_n$(say)

Integrating by parts,

$\displaystyle I_n=(\ln x)^n\int dx-\int\left(\frac{d (\ln x)^n}{dx}\int dx\right)dx$

$\displaystyle=x(\ln x)^n-n\int(\ln x)^{n-1}dx$

$\displaystyle\implies I_n=x(\ln x)^n-nI_{n-1} $

Now put $n=m-1$