# Integration by parts

1. Dec 20, 2007

### ricekrispie

little help? simple question...

1. The problem statement, all variables and given/known data

Integrate:
xlnx dx

use integration by parts

2. Relevant equations

3. The attempt at a solution

let u = x du = dx
dv = ln x v = ??

2. Dec 20, 2007

### rock.freak667

If you don't know the integral of lnx...try letting u=lnx and dx=x dx instead of the way you wrote it

3. Dec 20, 2007

### ricekrispie

1. The problem statement, all variables and given/known data

Integrate:
xlnx dx

2. Relevant equations

3. The attempt at a solution

let u = x du = dx
dv = ln x v = ??

4. Dec 20, 2007

### nicksauce

Let u = lnx
du = 1/x
dv = xdx
v = x^2 / 2

5. Dec 20, 2007

### ricekrispie

if I have to integrate it from 1 to 2 where are these plugged in?

6. Dec 20, 2007

### nicksauce

$$\int_a^b udv = uv|_a^b - \int_a^b vdu$$
Or in other words
$$\int_a^b udv = u(b)v(b) - u(a)v(a) - \int_a^b vdu$$

7. Dec 20, 2007

### ricekrispie

my attempt

i wonder if this worked...

$$\int_1^2 x lnx dx = lnx (x^2/2)|_1^2 - \int_1^2 (x^2/2)* (1/x) dx$$
$$= 2ln2 - ((1/4)(x^2))|_1^2$$

= 2ln2 - (5/4)
??

8. Dec 20, 2007

### HallsofIvy

Staff Emeritus
Almost! $(1/4)(x^2)$ when x= 2 is 1. when x= 1, it is 1/4 so the last part is -(1- 1/4)= -3/4, not -5/4.

9. Dec 20, 2007

### rocomath

check your evaluation for the last part

it might be me but you did the integration correctly, gj.

10. Dec 20, 2007

### nicksauce

Almost...

$$x^2|^2_1$$ Is 3 not 5.

11. Dec 20, 2007

### ricekrispie

hehe i noticed a lack of parenthesis... thanks for helping