How do you integrate $ln(x)^35$ ?

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Answer 1 · 2016-07-04T14:08:59

$int (log_e x)^n dx = (c_0+sum_{k=0}^{n-1}((-1)^{n+k} n!)/((k+1)!)(log_e x)^{k+1})x$

Making $y = log_e x$ and
$dy = dx/x$

substituting in $int (log_e x)^n dx$ we have

$int (log_e x)^n dx equiv int y^n e^y dy$

Now

$d/(dy)(y^n e^y) = ny^{n-1}e^y+y^n e^y$

so

$int y^n e^y dy = y^n e^y-n int y^{n-1} e^y dy + c_1$

or calling $I_n = int y^n e^y dy$ we have

$I_n = y^n e^y -n I_{n-1} + c_1$ or supposing $c_1=0$

$I_n+nI_{n-1} = y^n e^y$ Now making $J_n=e^yI_n$

$J_n+nJ_{n-1} = y^n$

$J_n = (-1)^n n!(c_0+sum_{k=0}^{n-1}((-1)^k)/((k+1)!)y^{k+1})$

then

$I_n = (c_0+sum_{k=0}^{n-1}((-1)^{n+k} n!)/((k+1)!)(log_e x)^{k+1})x$

so

$int (log_e x)^n dx = (c_0+sum_{k=0}^{n-1}((-1)^{n+k} n!)/((k+1)!)(log_e x)^{k+1})x$

How do you integrate ln(x)^35ln(x)^35?