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, \(\displaystyle dv \ does \ not \ equal \ x^{n}. \ it \ equals \ x^{n}dx\)