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!
Best Answer
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 $n=m-1$