# New to Integration (and new to forum)

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• Sep 9th 2009, 05:18 PM
PandaNomium
New to Integration (and new to forum)
Hi guys, calculus is not my strong point and I'm having a bit of difficulty trying to integrate these problems:

integrate: x(x-2)/(x-1)^3
int: x/(9-x)^1/2

Can I use u substitution here?

and also I'm not sure how to go about these problems, where I have to find the area under the curve bounded the given numbers:

1/xlnx where b=e^2 and a=e

x-1/(x+1) where b=1 and a=0

I would really appreciate it!

<3
from a humanities major
• Sep 9th 2009, 05:47 PM
Krizalid
The key is to exterminate the root, that is what annoys us, so for your first problem, put $\displaystyle t=\sqrt[3]{x-1}$ written backwards as $\displaystyle t^3=x-1,$ and then compute $\displaystyle \frac{dt}{dx}.$

In the same fashion, can you compute your second integral?
• Sep 9th 2009, 06:17 PM
PandaNomium
I'm not sure what you mean since my first problem isn't a root, it's cubed. am I supposed to take a cubed root there?
• Sep 9th 2009, 06:18 PM
PandaNomium
oh wait, do you mean x(x-2)^(1/3) / (x-1)
• Sep 9th 2009, 06:18 PM
Krizalid
Ohh!! yes, you're right! ahhh well, it's even simple! put $\displaystyle t=x-1$ and use that $\displaystyle \frac{a+b}c=\frac ac+\frac bc.$

anyway, what I did before still works for your second problem.
• Sep 9th 2009, 06:36 PM
PandaNomium
thanks for replying so quickly!

I suppose I don't understand what you mean by the a+b/c = a/c +b/c. can I still do that when x is multiplied by (x-2)? Maybe I'm missing a step here.

x(x-2) x x-2
----------- ?=? ---------- + ---------
(x-1)^3 (x-1)^3 (x-1)^3

i'm sorry if i'm a little bit slow. I'm taking calc 2, and haven't taken calculus since high school.
• Sep 9th 2009, 06:45 PM
Krizalid
Quote:

Originally Posted by PandaNomium

integrate: x(x-2)/(x-1)^3

Is this your integral right? so then put $\displaystyle t=x-1,$ what do you get?
• Sep 9th 2009, 06:48 PM
PandaNomium
(t+1)(t-1)/ t^3 ?
• Sep 9th 2009, 06:50 PM
Krizalid
yes, now that is $\displaystyle \frac{t^2-1}{t^3}=\frac1t-\frac1{t^3},$ and those are easy to integrate.
• Sep 9th 2009, 06:54 PM
PandaNomium
oooh.

is it ln(x-1) + 1/2(x-1)^2 + C?
• Sep 9th 2009, 06:55 PM
Krizalid
correct! (i see you added the constant! well done!)
• Sep 9th 2009, 07:03 PM
PandaNomium
thanks my brain is a little fuzzy... :P
• Sep 9th 2009, 07:24 PM
PandaNomium
so for the second problem I set t^2=9-x^2. Then when i derive it, it becomes t=x?

I'm kind of stuck there.
• Sep 9th 2009, 07:25 PM
Krizalid
no, it's just $\displaystyle t^2=9-x.$
• Sep 9th 2009, 07:28 PM
PandaNomium
oh i'm sorry, i wrote the problem wrong:

it's suppose to be x/sqrt(9-x^2)
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