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

Jul 4, 2016

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

#### Explanation:

Making $y = {\log}_{e} x$ and
$\mathrm{dy} = \frac{\mathrm{dx}}{x}$

substituting in $\int {\left({\log}_{e} x\right)}^{n} \mathrm{dx}$ we have

$\int {\left({\log}_{e} x\right)}^{n} \mathrm{dx} \equiv \int {y}^{n} {e}^{y} \mathrm{dy}$

Now

$\frac{d}{\mathrm{dy}} \left({y}^{n} {e}^{y}\right) = n {y}^{n - 1} {e}^{y} + {y}^{n} {e}^{y}$

so

$\int {y}^{n} {e}^{y} \mathrm{dy} = {y}^{n} {e}^{y} - n \int {y}^{n - 1} {e}^{y} \mathrm{dy} + {c}_{1}$

or calling ${I}_{n} = \int {y}^{n} {e}^{y} \mathrm{dy}$ we have

${I}_{n} = {y}^{n} {e}^{y} - n {I}_{n - 1} + {c}_{1}$ or supposing ${c}_{1} = 0$

${I}_{n} + n {I}_{n - 1} = {y}^{n} {e}^{y}$ Now making ${J}_{n} = {e}^{y} {I}_{n}$

${J}_{n} + n {J}_{n - 1} = {y}^{n}$

This recurrence equation has the solution
https://en.wikipedia.org/wiki/Recurrence_relation

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