For n any non-negative integer evaluate the integral :

$\int x^n \ln (x) dx$

Attempt to solution:

use integration by parts

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

$u=\ln (x)$

$du=1/x$

$\int u dv=\ln (x) \frac{x^{n-1}}{n-1}-\int\frac{x^{n-1}}{n-1} \frac{1}{x}$

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

2. Originally Posted by nyasha
For n any non-negative integer evaluate the integral :

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

Attempt to solution:

use integration by parts

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

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

I'm stuck here how do l further simplify this thing ?
$\int x^n \ln{x} \, dx$

$u = \ln{x}$ ... $du = \frac{1}{x} \, dx$

$dv = x^n \, dx$ ... $v = \frac{x^{n+1}}{n+1}$

$\int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{1}{n+1} \int x^n \, dx$

$\int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{x^{n+1}}{(n+1)^2} + C$

3. Hello, nyasha!

At this site, you must use [math ] [/math ] for LaTeX.

For any non-negative integer $n$, evaluate: . $\int x^n\ln(x)\,dx$
I would do it like this . . .

. . $\begin{array}{ccccccc}u &=& \ln x & & dv &=& x^n\,dx \\ du &=& \frac{dx}{x} & & v &=& \frac{x^{n+1}}{n+1} \end{array}$

Then we have: . $\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{n+1}\int x^n\,dx \;\;=\;\;\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{(n+1)^2}\,x^{n+1} + C$

. . . . . . $= \;\frac{1}{(n+1)^2}\,x^{n+1}\bigg[(n+1)\ln x - 1\bigg] + C$

Edit: Darn, too slow again!
.

4. Originally Posted by skeeter
$\int x^n \ln{x} \, dx$

$u = \ln{x}$ ... $du = \frac{1}{x} \, dx$

$dv = x^n \, dx$ ... $v = \frac{x^{n+1}}{n+1}$

$\int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{1}{n+1} \int x^n \, dx$

$\int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{x^{n+1}}{(n+1)^2} + C$

Thanks

5. Originally Posted by Soroban
Hello, nyasha!

At this site, you must use [math ] [/math ] for LaTeX.

I would do it like this . . .

. . $\begin{array}{ccccccc}u &=& \ln x & & dv &=& x^n\,dx \\ du &=& \frac{dx}{x} & & v &=& \frac{x^{n+1}}{n+1} \end{array}$

Then we have: . $\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{n+1}\int x^n\,dx \;\;=\;\;\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{(n+1)^2}\,x^{n+1} + C$

. . . . . . $= \;\frac{1}{(n+1)^2}\,x^{n+1}\bigg[(n+1)\ln x - 1\bigg] + C$

Edit: Darn, too slow again!
.

Thanks,actually that was the simplified format they wanted in the answer text.