Question

How do you integrate `int x^nlnx` by integration by parts method?

Answer 1

`int x^n log_e x dx = x^{n+1}(((n+1)log_e x-1)/(n+1)^2)`

Explanation:

`d/(dx)(x^{n+1}log_e x) = (n+1)x^nlog_ex+x^n`

then

`(n+1)int x^n log_e x dx = x^{n+1}log_e x - 1/(n+1)x^{n+1}`

so

`int x^n log_e x dx = x^{n+1}(((n+1)log_e x-1)/(n+1)^2)`