1. ## Integration by Part

So my professor was going over integration by part and using the tabular method. He use the problem:

(the integral sign, big swoosh like thing) (3t+9)^3(2t+e^2t)dt

Now, to find the derivative, I believe he used the chain rule, perhaps? Only thing is, I am not certain how I would use the chain rule on this one? Somehow, I do recall him showing on the outside he obtained

3x3(something) for the first derivative
and
3x2x9(something) for the second derivative

But I do not know how.

Any help would be appreciated!

2. Originally Posted by Mike5055
So my professor was going over integration by part and using the tabular method. He use the problem:

(the integral sign, big swoosh like thing) (3t+9)^3(2t+e^2t)dt

Now, to find the derivative, I believe he used the chain rule, perhaps? Only thing is, I am not certain how I would use the chain rule on this one? Somehow, I do recall him showing on the outside he obtained

3x3(something) for the first derivative
and
3x2x9(something) for the second derivative

But I do not know how.

Any help would be appreciated!
The derivative of which? Which did you select to be the "u" term?

3. I'm almost positive he set (3t+9)^3 = u

4. Originally Posted by Mike5055
I'm almost positive he set (3t+9)^3 = u
Ok then if we let $\displaystyle f(x)=(3x+9)^3$ then

$\displaystyle f'(x)=9(3t+9)^2$

$\displaystyle f''(x)=54(3x+9)$

$\displaystyle f'''(x)=162$

5. That was it!

Alright, my question is, how did you get the 9, the 54, and eventually, the 162?

6. Originally Posted by Mike5055
That was it!

Alright, my question is, how did you get the 9, the 54, and eventually, the 162?
$\displaystyle \left(3x+9\right)^3=\left(3(x+3)\right)^3=27(x+3)^ 3$

And then do you know the chain rule?

$\displaystyle \frac{d}{dx}\bigg[f\left(g(x)\right)\bigg]=f'\left(g(x)\right)\cdot{g'(x)}$

7. Originally Posted by Mathstud28
Ok then if we let $\displaystyle f(x)=(3x+9)^3$ then

$\displaystyle f'(x)=9(3t+9)^2$

$\displaystyle f''(x)=54(3x+9)$

$\displaystyle f'''(x)=162$
This is all correct. You want :
$\displaystyle f(x)=(3x+9)^3$
$\displaystyle f'(x)=3(3x+9)^2*3$
$\displaystyle f'(x)=9(3x+9)^2$

Do this in a similar manner until you get down to 162. Do you understand about using the tabular method and integrating $\displaystyle 2t+e^(2t)$?

8. Oh! Okay! I get it now. I was doing it wrong at first and wasn't getting anything near those. Thank you!

9. Originally Posted by kathrynmath
Do this in a similar manner until you get down to 162. Do you understand about using the tabular method and integrating $\displaystyle 2t+e^(2t)$?
Somewhat. My professor never really explains things very clearly so I have to rely on the textbook for most of it.

10. Originally Posted by Mike5055
Somewhat. My professor never really explains things very clearly so I have to rely on the textbook for most of it.
Ok, what helped me learn it was by making a chart with one colum labeled D(for taking the derivative) and the other I(for integrating).

So, I'll start you out:

D
$\displaystyle (3t+9)^3$
$\displaystyle 9(3t+9)^2$

I
$\displaystyle 2t+e^(2t)$
$\displaystyle 1/3t^3+1/2e^(2t)$

You understand that you once you get all the terms, you start by multiplying $\displaystyle (3t+9)^3$ ($\displaystyle 1/3t^3+1/2e^(2t)$)?

11. Yeah, the 2 column thing is what a friend taught me to do and it definitely helps a lot.