# Integration by Part

• Nov 8th 2008, 01:13 PM
Mike5055
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!
• Nov 8th 2008, 01:17 PM
Mathstud28
Quote:

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?
• Nov 8th 2008, 01:22 PM
Mike5055
I'm almost positive he set (3t+9)^3 = u
• Nov 8th 2008, 01:27 PM
Mathstud28
Quote:

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

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

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

$\displaystyle f'''(x)=162$
• Nov 8th 2008, 01:35 PM
Mike5055
That was it!

Alright, my question is, how did you get the 9, the 54, and eventually, the 162?
• Nov 8th 2008, 01:37 PM
Mathstud28
Quote:

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?

$\displaystyle \frac{d}{dx}\bigg[f\left(g(x)\right)\bigg]=f'\left(g(x)\right)\cdot{g'(x)}$
• Nov 8th 2008, 01:37 PM
kathrynmath
Quote:

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 $\displaystyle 2t+e^(2t)$?
• Nov 8th 2008, 01:37 PM
Mike5055
Oh! Okay! I get it now. I was doing it wrong at first and wasn't getting anything near those. Thank you!
• Nov 8th 2008, 01:39 PM
Mike5055
Quote:

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.
• Nov 8th 2008, 01:50 PM
kathrynmath
Quote:

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 $\displaystyle (3t+9)^3$ ($\displaystyle 1/3t^3+1/2e^(2t)$)?
• Nov 8th 2008, 01:51 PM
Mike5055
Yeah, the 2 column thing is what a friend taught me to do and it definitely helps a lot.