Question

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

Answer 1

`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
`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`

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`