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Tricky indefinite integral problem

  1. Jun 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Ok the problem is:
    ∫-1/(4x-x^2) dx

    The answer in the back of the book is:
    (1/4)ln(abs((x-4)/x))) + C

    2. Relevant equations
    I think this would be used somehow:
    ∫ du/(a^2-u^2) = 1/2a ln(abs((a+u)/(a-u))) + C


    3. The attempt at a solution
    ∫-1/(4x-x^2) dx

    ∫-1/(-x^2+4x) dx

    ∫-1/-(x^2-4x) dx

    ∫1/(x^2-4x) dx

    ∫1/(x^2-4x+4-4) dx

    ∫1/((x-2)^2-4) dx

    ∫1/((x-2)^2-2^2) dx

    At this point I'm stumped. No matter what I try I can seem to get it to look like the book answer.

    If I move the bottom part around and use u-substitution I get:

    ∫1/(u^2-2^2) du let u=x-2 and du = dx

    ∫1/(-2^2+u^2) du

    ∫1/-(2^2-u^2) du

    ∫-1/(2^2-u^2) du

    -∫1/(2^2-u^2) du

    I will get a negative number which isn't what I want. What am I doing wrong? I would greatly appreciate any help. Thanks :)
     
  2. jcsd
  3. Jun 26, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    Use partial fractions:
    [tex]
    \frac{-1}{4x-x^2} = \frac{1}{x(x-4)}
    [/tex]
    Now find A and B where
    [tex]
    \frac{1}{x(4-x)} \equiv \frac Ax + \frac B{x-4}
    [/tex]

    You should now be able to integrate each term separately.

    (You can write
    [tex]
    \frac{1}{x^2 - 4x} = \frac{1}{(x - 2)^2 - 2^2}
    [/tex]
    and substitute [itex]u = x - 2[/itex], but the partial fraction approach is the usual method.)
     
    Last edited: Jun 26, 2013
  4. Jun 26, 2013 #3

    Mark44

    Staff: Mentor

    Use partial fractions to decompose the integrand into A/(u - 2) + B/(u + 2).

    IOW, solve this equation for A and B:
    $$\frac{1}{(u - 2)(u + 2)} = \frac{A}{u - 2} +\frac{B}{u + 2} $$

    The equation above actually must be identically true (true for any values of u other than 2 and -2.
     
  5. Jun 26, 2013 #4
    Thanks for the help guys, but how would I go about doing it without partial fractions? I'm not supposed to use that technique yet. Sorry I forgot to mention that in my post.
     
  6. Jun 26, 2013 #5

    Mark44

    Staff: Mentor

    A trig substitution should work. Have you learned that technique yet?
     
  7. Jun 26, 2013 #6

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    I assume you got that from a table you are allowed to use?

    Try pasmith's suggestion ##u=x-2## and ##a=2## in your table formula above.
     
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