# Integration by Parts

1. May 26, 2005

### cogs24

hi guys
just doing some revision and im stuck on this question

*integral sign* x^2 . exponential ^ -3x . dx

I know i have to use integration by parts, but i just cant seem to get it out
any ideas?
thanx

2. May 26, 2005

### dextercioby

HINT:Take $$dv=e^{-3x} \ dx$$ and $$u=x^2$$.

U'll figure out what to do next.

Daniel.

3. May 26, 2005

### cogs24

but, what is the integral of exponential ^ -3x?

4. May 26, 2005

### cogs24

is it -1/3 *exponential* ^ -3x?

5. May 26, 2005

### dextercioby

Yes,as you can check by differentiation.

Daniel.

6. May 26, 2005

### Jameson

The tabular method would work great on this question. Are you familiar with this?

7. May 26, 2005

### cogs24

ok, once i sub u, du, v and dv into the integral by parts formula, i have to assign, u and dv again to:
-1/3 exponential ^ -3x and 2x

so is dv assigned to the exponential again, like in the first case?

8. May 26, 2005

### cogs24

no, i am not aware of the tabular method, sorry

9. May 26, 2005

### cogs24

ok, ive got my answer, unfortunately i dont kow if its right, we dont get solutions for this exercise

-1/3 *exponential* ^ -3x . x^2 - 1/9*exponential*^-3x . 2x - 2/27*exponential*^-3x

10. May 26, 2005

### whozum

Maple agrees, good job.

11. May 26, 2005

### dextercioby

Daniel.

12. May 27, 2005

### cogs24

arbitary constant would be + c

13. May 27, 2005

### Jameson

In case you were interested in this method, here is a link to the explanation. If one of the two terms will eventually differentiate to zero, this method is much less time consuming.

Jameson

14. May 28, 2005

### cogs24

i see, thanx for that link, i will probably apply this method from now on