# Thread: New to Integration (and new to forum)

1. ## 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

2. 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?

3. 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?

4. oh wait, do you mean x(x-2)^(1/3) / (x-1)

5. 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.

6. 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.

7. 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?

8. (t+1)(t-1)/ t^3 ?

9. yes, now that is $\displaystyle \frac{t^2-1}{t^3}=\frac1t-\frac1{t^3},$ and those are easy to integrate.

10. oooh.

is it ln(x-1) + 1/2(x-1)^2 + C?

11. correct! (i see you added the constant! well done!)

12. thanks my brain is a little fuzzy... :P

13. 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.

14. no, it's just $\displaystyle t^2=9-x.$

15. oh i'm sorry, i wrote the problem wrong:

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

Page 1 of 2 12 Last