# Integration (Sort of)

• Feb 16th 2010, 12:31 AM
Awsom Guy
Integration (Sort of)
Find F(x) given f(x) is:
(x+2)(x^2+ 4x + 3)^5
Well my problem is what do I do with the brackets. I tried but I can't get the answer right.
My attempt:
u=(x^2 + 4x + 3)
u'=2x+4
u^5
u^6/6 + C
(x^2 + 4x + 3)^6/6 + C
I should have 12 instead of 6 on the bottom.
(x^2 + 4x + 3)^6/12 + C
Thank you
• Feb 16th 2010, 12:56 AM
mathemagister
Quote:

Originally Posted by Awsom Guy
Find F(x) given f(x) is:
(x+2)(x^2+ 4x + 3)^5
Well my problem is what do I do with the brackets. I tried but I can't get the answer right.
My attempt:
u=(x^2 + 4x + 3)
u'=2x+4
u^5
u^6/6 + C
(x^2 + 4x + 3)^6/6 + C
I should have 12 instead of 6 on the bottom.
(x^2 + 4x + 3)^6/12 + C
Thank you

You're forgetting the chain rule. Remember when you take take the derivative of that, you need to multiply it by the derivative of the inside. So, compensate for that when integrating by multiplying it by 1/(derivative of inside). The key here is noticing that the derivative of the inside (2x+4) is 2(x+2), notice the x+2 on the outside and you're all set. Feel free to ask me for any more help? (Happy)
• Feb 16th 2010, 01:01 AM
Awsom Guy
Is that for integration.
• Feb 16th 2010, 01:05 AM
mathemagister
Quote:

Originally Posted by Awsom Guy
Is that for integration.

Since integration is just the opposite of differentiation, you need to do the opposite of the chain rule: instead of multiplying by the derivative of the inside, you need to divide by it. You can always check your answer by differentiating your answer and see if you get the question.
• Feb 16th 2010, 01:09 AM
Awsom Guy
I know that :) but how do I do it properly wait I'll try and then give you a buzz. :)
• Feb 16th 2010, 01:15 AM
Prove It
Quote:

Originally Posted by Awsom Guy
Find F(x) given f(x) is:
(x+2)(x^2+ 4x + 3)^5
Well my problem is what do I do with the brackets. I tried but I can't get the answer right.
My attempt:
u=(x^2 + 4x + 3)
u'=2x+4
u^5
u^6/6 + C
(x^2 + 4x + 3)^6/6 + C
I should have 12 instead of 6 on the bottom.
(x^2 + 4x + 3)^6/12 + C
Thank you

$\int{(x + 2)(x^2 + 4x + 3)^5\,dx} = \frac{1}{2}\int{2(x + 2)(x^2 + 4x + 3)^5\,dx}$.

Let $u = x^2 + 4x + 3$ so that $\frac{du}{dx} = 2x + 4 = 2(x + 2)$

So the integral becomes

$\frac{1}{2}\int{u^5\,\frac{du}{dx}\,dx}$

$= \frac{1}{2}\int{u^5\,du}$

$= \frac{1}{2}\left[\frac{1}{6}u^6\right] + C$

$= \frac{1}{12}u^6 + C$

$= \frac{1}{12}(x^2 + 4x + 3) + C$.
• Feb 16th 2010, 01:17 AM
mathemagister
Quote:

Originally Posted by Awsom Guy
I know that :) but how do I do it properly wait I'll try and then give you a buzz. :)

Remember the key is noticing that the derivative of the inside is (2x+4) which is exactly 2 times too big. (the outside is x+2)
• Feb 16th 2010, 01:19 AM
mathemagister
Quote:

Originally Posted by Prove It
$\int{(x + 2)(x^2 + 4x + 3)^5\,dx} = \frac{1}{2}\int{2(x + 2)(x^2 + 4x + 3)^5\,dx}$.

Let $u = x^2 + 4x + 3$ so that $\frac{du}{dx} = 2x + 4 = 2(x + 2)$

So the integral becomes

$\frac{1}{2}\int{u^5\,\frac{du}{dx}\,dx}$

$= \frac{1}{2}\int{u^5\,du}$

$= \frac{1}{2}\left[\frac{1}{6}u^6\right] + C$

$= \frac{1}{12}u^6 + C$

$= \frac{1}{12}(x^2 + 4x + 3) + C$.

Don't ruin it for him. At least put a spoiler around.

This is part of the MHF terms of agreement:
"When you see that someone is already helping a member and is waiting for the OP to give feedback, please don't give a full solution. "
• Feb 16th 2010, 01:21 AM
HallsofIvy
Quote:

Originally Posted by Awsom Guy
Find F(x) given f(x) is:
(x+2)(x^2+ 4x + 3)^5
Well my problem is what do I do with the brackets. I tried but I can't get the answer right.
My attempt:
u=(x^2 + 4x + 3)
u'=2x+4
u^5
u^6/6 + C
(x^2 + 4x + 3)^6/6 + C

Part of your problem is that you don't work "cleanly". For example, you say "u'= 2x+4" but you don't say why you need to know that or what you are doing with it. The very next line is "u^5" which is meaningless. I suspect that if you had written out the entire integral you would have seen where you went wrong.

u= x^2+ 4x+ 3 so u'= du/dx= 2x+ 4 and du= (2x+4)dx. The whole point of taking the derivative there is to be able to decide how to replace "dx" in the integral. The crucial point is that du= (2x+4)dx but your integral has, in addition to (x^2+ 4x+ 3)^5, (x+2)dx, not (2x+4)dx.

There are two paths people use. From your du= (2x+4)dx, you can factor a "2" out and have du= 2(x+2)dx so du/2= (x+2)dx.

Or you can write dx= du/(2x+4) and replace that: (x+2)dx= (x+2)/(2x+4) du= (x+2)/(2(x+2)) du= (1/2)du.

Either way, your integral is now $\frac{1}{2}\int u^5 du$, not just $\int u^5 du$.

Quote:

I should have 12 instead of 6 on the bottom.
(x^2 + 4x + 3)^6/12 + C
Thank you
• Feb 16th 2010, 01:22 AM
Awsom Guy
Thanks heaps prove it, this is how I learn. :)
• Feb 16th 2010, 01:25 AM
Prove It
Quote:

Originally Posted by mathemagister
Don't ruin it for him. At least put a spoiler around.

This is part of the MHF terms of agreement:
"When you see that someone is already helping a member and is waiting for the OP to give feedback, please don't give a full solution. "

Some people, like the OP for example, learn by having a solution to refer back to to check the answer they have come up with. The OP said he would be back after trying it himself.
• Feb 16th 2010, 01:25 AM
Awsom Guy
Thanks guys, I realised my problem I forgot the half in the front. Or else i would have gottn it :) thanks guys.
• Feb 16th 2010, 01:31 AM
mathemagister
Quote:

Originally Posted by Prove It
Some people, like the OP for example, learn by having a solution to refer back to to check the answer they have come up with. The OP said he would be back after trying it himself.

I know, but I was just saying, because that got me a warning before.

And Awsom Guy, great that you understand it now (Happy)
• Feb 16th 2010, 01:32 AM
Awsom Guy
Oh, thanks anyways I knew I needed a half in there somehow but i wasn;t sure how to get it thanks everyone. :)