Evaluating a complicated integral: int x^n Ln(x) dx

[chengeto]

For n any non-negative integer evaluate the integral :

$\displaystyle \int x^n Ln(x) dx$

Attempt to solution:

use integration by parts

$\displaystyle dv=x^n$
$\displaystyle v=\frac{x^{n-1}}{n-1}$

$\displaystyle u=Ln(x)$
$\displaystyle du=1/x$
\(\displaystyle \int udv=Ln(x)\frac{x^{n-1}}{n-1}-\int\frac{x^{n-1}}{n-1}\frac{1}x}\)

I'm stuck here how do l further simplify this thing ?

 

[Deleted member 4993]

For n any non-negative integer evaluate the integral :

$\displaystyle \int x^n In(x) dx$


Attempt to solution:

use integration by parts

$\displaystyle dv=x^n$
$\displaystyle v=\frac{x^{n-1}}{n-1}$

$\displaystyle u=In(x)$
$\displaystyle du=1/x$
\(\displaystyle \int udv=In(x)\frac{x^{n-1}}{n-1}-\int\frac{x^{n-1}}{n-1}\frac{1}x}\)

I'm stuck here how do l further simplify this thing ?

$\displaystyle \frac{x^{n-1}}{x} \, = \, x^{n-2}$

change 'In' to 'Ln' in your post to avoid confusion.

 

[chengeto]

For n any non-negative integer evaluate the integral :

$\displaystyle \int x^n In(x) dx$


Attempt to solution:

use integration by parts

$\displaystyle dv=x^n$
$\displaystyle v=\frac{x^{n-1}}{n-1}$

$\displaystyle u=In(x)$
$\displaystyle du=1/x$
\(\displaystyle \int udv=In(x)\frac{x^{n-1}}{n-1}-\int\frac{x^{n-1}}{n-1}\frac{1}x}\)

I'm stuck here how do l further simplify this thing ?

$\displaystyle \frac{x^{n-1}}{x} \, = \, x^{n-2}$

change 'In' to 'Ln' in your post to avoid confusion.

Thanks for the help

 

[BigGlenntheHeavy]

$\displaystyle dv = x^{n}dx$

$\displaystyle \int{dv} = \int{x^{n}}dx$

$\displaystyle v = \frac{x^{n+1}}{n+1},\ not\ \frac{x^{n-1}}{n-1}$

Also, to be picky, $\displaystyle dv \ does \ not \ equal \ x^{n}. \ it \ equals \ x^{n}dx$