1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrating 1/(1-x)^2 its making me crazy

  1. Apr 22, 2015 #1
    • Member informed about compulsory use of homework template.
    1. The problem statement, all variables and given/known data

    I know its easy, but I'm making a mistake somewhere that is making me crazy. I want to solve \begin{equation} y = \int 1/(1-x)^2 \cdot dx \end{equation}
    I use de sixth formula in this PDF, but it does not work http://integral-table.com/downloads/single-page-integral-table.pdf

    2. Relevant equations

    \begin{equation} y = \int 1/(1-x)^2 \cdot dx \end{equation}

    3. The attempt at a solution

    This is what I do.
    \begin{equation} y = \int 1/(1-x)^2 \cdot dx = \int (1-x)^{-2} \cdot dx = \frac{(1-x)^{-1}}{-1} = \frac{1}{x-1} \end{equation}

    What is wrong?
     
    Last edited: Apr 22, 2015
  2. jcsd
  3. Apr 22, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, you haven't shown your work, so how can anyone say what is wrong?
     
  4. Apr 22, 2015 #3
    Sorry. This is what I do.
    ∫1/(1-x)^2dx = ∫(1-x)^-2dx = (1-x)^-1/-1 = 1/(x-1)
     
  5. Apr 22, 2015 #4

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The formula you're trying to use says a+x, not a-x, so you should start with a rewrite that puts a plus sign in front of the x.
     
  6. Apr 22, 2015 #5
    But that formula is valid for any other exponent I think.

    Look at this (this is correct).

    \begin{equation} y = \int 1/(1-x)^5 \cdot dx = \int (1-x)^{-5} \cdot dx = \frac{(1-x)^{-4}}{-4} = \frac{1}{4(x-1)^4} \end{equation}

    That's why I am confused.
     
  7. Apr 22, 2015 #6
    Just a question from me: how do you enter an equation?

    And notice that (1-x)^(-4)/(-4) does not equal to the two terms beside it. Basically you did two mistakes that made the sign change twice. [(1-x)^-4 = (x-1)^-4]
     
  8. Apr 22, 2015 #7

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Choose "Help/How-To" on the "Info" menu in the upper right. Then click on "LaTeX Primer".
     
  9. Apr 22, 2015 #8
    I do this -4*(1-x)^-4 = 4*(x-1)^-4, that's why I change the sign.
     
  10. Apr 22, 2015 #9

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    IYour first post is nearly right except there is a simple error in the last step, you have confused rules about straight minuses and minuses in an index it seems.
     
  11. Apr 22, 2015 #10
    (1-x)^-4 = (-(x-1))^-4 = (-1)^-4 * (x-1)^-4 = (x-1)^-4
     
  12. Apr 22, 2015 #11

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The last step is correct. It's the one before that (the one where he uses the formula from the pdf) that's wrong.
     
  13. Apr 22, 2015 #12
    ¿Do you mean that?

    \begin{equation} y = -\int -1/(1-x)^2 \cdot dx = -\int -(1-x)^{-2} \cdot dx = -\frac{(1-x)^{-1}}{-1} = \frac{1}{1-x} \end{equation}

    But if I change the exponent to 5, as in the other post, and I do the same, I don't get the correct one.

    \begin{equation} y = -\int -1/(1-x)^5 \cdot dx = -\int -(1-x)^{-5} \cdot dx = -\frac{(1-x)^{-4}}{-4} = \frac{1}{4*(1-x)^4} \end{equation}


    I'm missed up :(
     
  14. Apr 22, 2015 #13

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Those extra two minus signs in the first calculation don't help. What I wanted you to do is to find a way to rewrite ##\frac{1}{(1-x)^2}## in the form ##\frac{a}{(b+x)^2}##. You need a plus sign directly in front of the x before you can apply the formula.

    Edit: My first version of this post contained an additional statement that was wrong. If you saw it, just ignore it. You should do a rewrite of the type described above at the start of both of these calculations. If you do, the results should be OK.
     
    Last edited: Apr 22, 2015
  15. Apr 22, 2015 #14

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I interpreted this as a question about the calculations in post #12, but since you were quoting me, I guess you were asking if I'm sure that the comment I had made earlier (the one you quoted) is right. I have checked it again, and I still say that you're using formula (6) wrong in post #1.

    I will rephrase my comment about the rewrite you should do. If you want to be able to find primitive functions of ##\frac{1}{(1-x)^n}## for arbitrary n, you should rewrite ##\int\frac{1}{(1-x)^n}dx## in the form ##b\int(x+a)^m dx## and then use formula (6) to find ##\int(x+a)^m dx##. In fact, I think you should leave n arbitrary (except for the requirement ##n\neq 1##) and try this exact thing.

    What you did in post #1 was to assume that you would end up with the right-hand side of formula (6) even though what you had didn't match the left-hand side. What you did in post #12 was to assume that the answer would be wrong by a factor of -1 regardless of what the exponent is. Both of these assumptions are wrong.
     
    Last edited: Apr 22, 2015
  16. Apr 22, 2015 #15

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I would add that, unless the OP was required to use a table for this problem for some reason, he shouldn't use the table in the first place. Just do a substitution ##u=1-x##. He might even have gotten the correct answer the first time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integrating 1/(1-x)^2 its making me crazy
  1. 1/x^2 integral (Replies: 4)

Loading...