# Flawed Knowledge of Reduction Formulas

• Jan 26th 2010, 06:12 PM
LithiumPython
Flawed Knowledge of Reduction Formulas
Any explanation of how to evaluate this problem through the use of reduction formulas would be greatly appreciated.

Evaluate: $\displaystyle \int(x^nln(x)dx)$ where $\displaystyle x\not=-1$

I know that one must use integration by parts ($\displaystyle \int(udv)=uv-\int vdu$)
As it stands right now I have $\displaystyle u=x^n$ $\displaystyle \rightarrow du=nx^{n-1}dx$
Along with $\displaystyle dv=ln(x)dx$ $\displaystyle \rightarrow v=xln(x)-x+C$

I substitute the above values into the equation for integration by parts and I do not know how to continue from there. Any advice would be appreciated.
• Jan 26th 2010, 06:23 PM
tonio
Quote:

Originally Posted by LithiumPython
Any explanation of how to evaluate this problem through the use of reduction formulas would be greatly appreciated.

Evaluate: $\displaystyle \int(x^nln(x)dx)$ where $\displaystyle x\not=-1$

I know that one must use integration by parts ($\displaystyle \int(udv)=uv-\int vdu$)
As it stands right now I have $\displaystyle u=x^n$ $\displaystyle \rightarrow du=nx^{n-1}dx$
Along with $\displaystyle dv=ln(x)dx$ $\displaystyle \rightarrow v=xln(x)-x+C$

I substitute the above values into the equation for integration by parts and I do not know how to continue from there. Any advice would be appreciated.

It looks pretty straightforward and what you did is right (without the integration constant, of course) , so:

$\displaystyle \int x^n\ln x\,dx= x^{n+1}\ln x - x^{n+1}-n\!\!\int x^n\ln x\,dx+n\!\!\int x^n\,dx\,\Longrightarrow$ $\displaystyle (n+1)\!\!\int x^n\ln x\,dx=x^{n+1}\left(\ln x-1+\frac{n}{n+1}\right) + C$ ...

Tonio